not converting the int into a string on if statement

Your code looks mostly correct, including your usage of parseInt, but perhaps you are mixing your return statements with console.log statements?

var waitTime = parseInt("53");    
if (waitTime > 20) {    
    console.log('a lot of time'); // use console.log rather than return
} else if (waitTime < 20) {    
    console.log('we can wait'); // use console.log rather than return
}

console.log(waitTime);

http://jsbin.com/kebagodura/edit?js,console,output

Alternatively (to Jonathan's answer), put the code in a function and then the returns would make sense. An improved version would see you passing in the wait time as an argument to the function.

function doIt(number) {

  // Don't forget the radix on parseInt
  var waitTime  = parseInt(number, 10);
  if (waitTime > 20) {
    return 'a lot of time'
  } else if (waitTime < 20) {
    return 'we can wait'
  }
}

var result = doIt('53'); // a lot of time
var result = doIt('12'); // we can wait

DEMO

You could also do it this way- instead of "return" you can do this:

var waitTime  = parseInt("53");
if (waitTime > 20) {
  waitTime = 'a lot of time';
} else if (waitTime < 20) {
  waitTime = 'we can wait';
}

console.log(waitTime);

You simply can do that this way, reassigning the variable values in the if/else clause:

var waitTime = parseInt("53");

if (waitTime > 20) {
   waitTime = 'a lot of time';
} else if (waitTime <= 20) {
   waitTime = 'we can wait';
}

console.log(waitTime);

You have an error in your code.

return 'a lot of time'

It is trying to return the text a lot of time from a function. but you never created a function so it will throw an error.

Closer to what you want:

function wait(waitTime) {
  if (waitTime > 20) {
    return 'a lot of time'
  } else if (waitTime < 20) {
    return 'we can wait'
  }
}
console.log(wait(YOUR_WAIT_TIME_HERE));

EDIT You don't need the parseInt function at all. (ie: '25' > 20) implicitly casts '25' to a Number.