If two paintings can be rotated to match each other, are they the same? Or do we fix the dodecahedron in place and then paint it? I'll solve the problem with the latter assumption, since it's much easier in that case. Then, let's label the faces $1$ to $12$ in such a way that faces $i$ and $i+1$ share an edge for each $i = 1, 2, \cdots, 11.$ Color each face white/black depending on the parity of the number of adjacent red edges to that face, with white meaning even. Notice that no matter which edges we choose, there will always be an even number of faces with an odd number of adjacent red edges. In this way, there are only $2^{11}$ possible white/black colorings of the faces. There are $2^{30}$ ways to either color or not color each red edge. We will claim that each of the $2^{11}$ possible ways is attained equally often. Consider the edges $e_1, e_2, \cdots, e_{11}$ where $e_i$ connects faces $i$ and $i+1$. We claim that the $2^{11}$ different subsets of the $e_i$'s each result in a different white/black coloring of the faces. This would clearly suffice, since we're basically working in $F_2$ and things "add up." To see that this is the case, notice that if two different sets of the $e_i$'s did equal, then taking their "X-OR" would give a nonempty set of $e_i$'s that resulted in a all-white coloring. However, it's easy to see that this is not the case, e.g. by considering the $e_i$ in that set with the smallest $i$. $\square$