A new game, but with colors

I think that SquareFinder's answer was not easy to follow, since it didn't explained many of the intermediate steps (it is still correct and is a good answer nonetheless).

Further after the part that you guess wrongly, you could end with a contradiction somewhere else very different than what is stated in that answer, depending on what you guessed, where you guessed it and what was your line of thought from that place onwards.

So, I decided to add my own new answer.

Step 1

First, let's use the standard sudoku technique of filling pencil marks:

Step 2

Obviously you can...

Eliminate "bad neighbours", i.e. those that would make two adjacent squares with the same color.

Hence:

Step 3

However, we can...

Remove candidates from rows, columns and regions where the two colors are already present. I.E. Those candidates of some color in rows, columns or regions that already have two squares of that same color.

These pink cells:

Hence:

Step 4

Oh, we have...

A naked single red candidate and two hidden single green candidates.

If you don't find those, they are the pink highlighted cells. And choosing them removes the grayed-out candidates in the cyan highlighted cells:

So:

Step 5

And once again...

Another naked red candidate (3rd row) and also a hidden single green in bottom left.

Once again, let's highlight those in pink and the affected cells in cyan with the affected candidates to be removed grayed-out:

So:

Step 6

Look at...

The green candidate on 2nd row, 7th column.

What it gives:

Choosing the green on the orange square eliminates the green from the pink squares and makes the bottommost right region with only one green. Hence, the green in the orange square is impossible.

So:

Step 7

Now let's use a bit more advanced sudoku technique:

X-Wing

How?

The greens in 1st and 4th row and column.

See it?

Hence...

Choosing a green on one of the pink cells force the green on the other pink cell and eliminates them from the cyan and orange cells. Choosing a green from one of the cyan cells force the green on the other cyan cell and eliminated them from the pink and orange cells.

Either way, the green on the orange cell is impossible. Choosing it means that at least one of those three sets of squares would be screwed: the 1st row; the 1st column; the 4th column.

So:

Step 8

Now going to the bottom...

Region-row intersection. Similar to standard's sudoku box-row intersection.

How?

The last two rows are lacking a blue square each one. However, the bottom middle region also needs blue squares. The remaing blue square in the 8th row must be in this region (orange-colored squares), so that region must have its second blue square in the 7th row (pink-colored squares) which would also be the second blue in the 7th row. This means that blue is impossible in the remaining squares of the 7th row (cyan-colored):

So:

Step 9 - The guess

Now, it becomes difficult to progress only by inferences. So, we do a guess:

Let's guess that the square at 7th row and 7th column is green!

So:

Proceeding by...

Removing bad neighbours and promoting singles until we can't do this anymore.

We do:

• Remove green bad neighbour in cell A, making yellow the single candidate.

• A induces a bad yellow neighbour in B. Removing it, red becomes a naked single candidate.

• A induces C having the only (hidden) single green in the row.

• C induces D having the only hidden single blue in the row.

• A induces E to be the remaining yellow in the line.

So (pink cells):

Proceeding:

• E induces F to not be yellow (it would be a bad neighbour). C also induces F to not be green (the region would already have two greens). So, F is red.

• F induces G, H and I to not have red as it would be a bad neighbour of F.

• The bottom right region already has two yellows (E and I), so J is induced (by I) to be red.

• The bottom right region is almost complete, lacking only one square. So, K should be blue (induced by J).

• We could still go on, but let's stop here for now.

And then we have that (cyan cells):

Hmmm, however...

The third column must have two reds. But it only have one and there is no other red candidate left there!

This means that:

Our guess was wrong! The 7th square of the 7th row isn't green!

So:

We come back to the place we made the guess and remove that bad green:

Step 10

We can easily proceed by...

Filling the chain of single green candidates.

That is:

All of those cells with a green candidate that weren't in that X-Wing.

So:

Step 11

Should be obvious:

We have a naked single candidate blue in the 3rd column and this makes yellow below it.

Then:

Step 12

Now...

There is a region with two yellows.

Hence...

We can eliminate a few yellow candidates (cyan squares) and this makes another be a hidden single (pink square):

So:

Step 13

Let's see...

The reds in those first rows.

What is up?

In the 1st row, all the red candidates are in the same A region (pink squares), and only one of them is true. This means that the second red square in this region must be only in the 2nd row (cyan cells). The second red in the 2nd row then must be one of the last two cells (lime cells) because those are in another B region. The 3rd row must also have a second red in the B region (orange cells). This means that in the 4th row, we can't have red in the B region (brown cell), hence, that cell is blue.

So:

Step 14

Now...

Bad neighbours!

That:

A has a blue bad neighbour, so it is red. A induces B, C and D to not be red (so they are yellow, green and blue, respectively). D induces E to not be blue.

So:

Step 15

Now it is time for...

Finishing up the greens!

This means:

Those pink cells (remember the X-Wing):

So:

Step 16

Note that...

The rightmost region intercepting the 2nd row already have two yellows, so can't have any more.

So:

Step 17

Let's do something similar to what we've done at...

Step 13.

That:

The 1st row must have two blues and span two regions, each of those already have one blue. So, the blues in the 1st row are not in the same region, so, one of them is in the cyan cells and the other is in the pink cells. This means that the orange cell can't be blue because it would be the third in the same region.

So:

Step 18

We have...

A hidden single blue candidate in the 6th column (pink cell). Choosing it also eliminates blue in the 3rd column (cyan cell).

So:

Step 19

Once more:

Region-row intersection.

That one:

The 2nd row must have a blue in the pink cells, which is also the remaining blue in that region. So the cyan cell can't be blue.

So:

Step 20

Now it is obvious...

The naked single red.

And also...

A very long stream of singles and bad neighbours from that (20 cells in total).

That:

The pink cells. Starts at A, which induces removal of candidates in B and C. B induces D. C induces E. E induces F and G. F induces H. G induces I and J. J induces K and L. K induces M and N. N induces O. B and O induces P. D and P induces Q. Q induces R. R induces S. S induces T.

Thus:

Step 21

That last step was a long stream, eh? Well...

We can still continue!

Can you see...

The two hidden single blues? In the 3rd column and the 6th row?

This means...

Another stream (but this one is shorter). Starts in A and B. Both induces C. C induces D. D induces E. E induces F.

So:

And we finished off the blues! We are almost done!

Step 22 - Finish him

To finish it, do you see that...

There are 2 columns that just lack they last square.

Thus:

A and B are yellow. A induces C. C induces D. D induces E. E induces F and G. We finish at H.

And we are done!