The answer is (D). Let us first count in how many ways we can color the cube so that opposite faces have different colors. We have $4 \cdot 3$ choices for faces 1 and 6, and as many for each of the pairs of faces 2 and 5 and 3 and 4. Overall we have $4^{3} \cdot 3^{3}$ colorings.

We must now subtract from these all the colorings that do not satisfy the first requirement, that is, all those in which we have used only 2 colors. The number of such colorings equals the number of ways of choosing 2 distinct colors among the 4 available, 6, multiplied by the number of possible colorings of the cube with those colors, $2^{3}$, since for each pair of faces we have 2 ways of using the two available colors.

The number of admissible colorings is therefore $4^{3} \cdot 3^{3} - 6 \cdot 8$, which is equal to $2^{4} \cdot 3 \cdot 5 \cdot 7$.