Judging from the exception that you get, removeAdjDuplicates returns an empty string, invalidating all indexes past zero.

Although your code performs length checking at the top, it also performs this assignment when the two initial characters are different:

s = s.charAt(0) + removeAdjDuplicates(s.substring(1));

This means that s can become a one-character string if removeAdjDuplicates returns an empty string.

I think the source of the error is sufficiently explained by @dasblinkenlight's answer.

Although not clearly stated in the question, it looks like you're trying to remove adjacent duplicate letters recursively (one of your comments mentions that you would expect output s for input geegs).

Here's an alternative way to do it:

    while(!s.equals(s = s.replaceAll("(.)\\1", "")));

It uses a regular expression to match and remove duplicate characters, and the while loop keeps executing this until the string is no longer being modified by the operation.

You should simplify your code:

public static String removeAdjDuplicates(String s) {
    if (s == null || s.length() < 2)
        return s;

    if (s.charAt(0) != s.charAt(1))
        return s.charAt(0) + removeAdjDuplicates(s.substring(1));

    return removeAdjDuplicates(s.substring(2));
}

Changes

• The first two if statements do the same thing (return s;) and can be combined into one. Some of the conditions are redundant and can be eliminated.
• The third if statement should immediately return instead of continuing into the fourth if statement (or you can instead change the fourth if statement into an else), because removedAdjDuplicates can return an empty String making s a length-one String when the fourth if is expecting at least a length-two String.
• The fourth if can be eliminated because if (s.charAt(0) != s.charAt(1)) failed in the third if, then the only alternative is that (s.charAt(0) == s.charAt(1)), so the check for that isn't necessary.