Yep, that will work. You approached this a little differently than I did, but it's the same idea:
The trick here is that there is necessarily an equal number of right and wrong answers, but you can set things so that the wrong answers are all clumped together under the same few hat arrangements while the right answers are spread out. The general idea for the solution then is to select some number of "bad" arrangements, where all the other ("good") arrangements differ by 1 hat from at least one of the bad sets, with the rule "If you see that the other three have hats corresponding to one of the bad arrangements, guess the opposite color from what your hat is in that arrangement", which leads to everyone guessing wrong on those, while at least one guesses right on all the others.
We have a general upper bound of n/(n+1) - 4/5 in this case - because the total number of right and wrong answers must be the same. For example, we can't survive 13 of the cases here, because we would need 13 right answers and 13 wrong answers, but with 3 bad cases left there are only 3*4 wrong answers available.
The simplest way to do it with 4 - and this is where the hint came in - is to ignore one of the people entirely and just consider the other three, because with 3 you can also achieve a 3/4 success rate. (Apparently, in the more general problem you can reach the upper bound of n/(n+1) if n = 2^k-1 for some k, by using the corresponding Hamming code.)
Rule: Ignore the fourth person. Among the remaining three, if anyone sees two like hats, guess the opposite.
(This is basically ghostofmars' rule 2a, but applied to both halves of the problem by ignoring A entirely.)