It is most certainly a group. In fact, it is isomorphic to a subgroup of the symmetric group $S_{54}$ generated by $6$ elements, corresponding to rotating each row to the right and each column up. To see this, one need only label the $54$ colored stickers on a cube with the numbers $1$ through $54$. Since each of the motions of the cube are permutations of the $54$ stickers, numbering the stickers gives us a subgroup of $S_{54}$. One such numbering gives
$ (1\; 10)(2\; 11)(3\; 12)(10\; 30)(11\; 29)(12\; 28)(30\; 39)(29\; 38)(28\; 37)(39\; 1)(38\; 2)(37\; 1) \cong \text{ rotating the top row right}$ and similarly the other $5$ elements are products of $12$ $2$-cycles. It is worth noting that each of these permutations is even, which gives an easy proof of the unsolvability of certain configurations, such as any configuration in which two stickers have been switched.
Edit: Some people consider the cube to be the same after being rotated as a whole. We can modify our group to deal with this by omitting the last row and column rotations, giving us a subgroup of $S_{54}$ generated by $4$ elements. We can do this because rotating the last row to the right is the same as rotating the cube to the right and the first two rows to the left, and a similar procedure works for columns.