I don't think this is solved by Burnside's Lemma since there is a condition that each side is painted a different colour. The question is as follows.
If I had a cube and six colours, and painted each side a different colour, how many (different) ways could I paint the cube? What about if I had $n$ colours instead of 6?
The answer given in an old thread on a different site is $6!$ for the first question, and $n(n-1)(n-2)(n-3)(n-4)(n-5)$ for the second question. However, this doesn't actually hold up because a few of the paintings are isomorphic. The original thread assumes we can somehow tell the difference between two paintings which actually look identical if you rotate the cube, which I don't think is what the question intended.
The answer I got for the first question is $4! + 4 = 28$. But this was just through a case-bash, and I'm not sure whether it's correct or whether it generalizes.