Solution:

There are nine non-equivalent colorings where two edges are red, two are black, and two are green. In the illustrations below, we will indicate the three colors by drawing edges thick, thin, or dashed. We will partition the colorings into three cases, determined by the number of pairs of edges that are colored the same: three, one, or zero (two is impossible). For each case, without loss of generality, we fix the position of the two thick edges and carefully examine all possible colorings, and determine which are equivalent.

- Each pair of opposite edges is colored the same. Fixing the two thick edges, there are two possible configurations shown below.

These two colorings are non-equivalent. To see why, note that the only rotations available are 2-fold (180-degree) rotations about axes joining midpoints of opposite sides, or 3-fold (120-degree) rotations about axes joining a vertex with the center of the opposite face. The 2-fold rotations leave both pictures alone (none of them turns one picture into the other), and the 3-fold rotations do not keep the thick lines in place.

- Exactly one pair of opposite sides is the same color. For example, suppose that the two opposite sides are colored with thick lines. Keeping this pair in place, there are only two configurations.

Note that these two colorings are in fact equivalent, since you can turn one into the other by performing a 2-fold rotation about the axis joining the midpoint of the two thick edges. Since there are three choices of colors we can use for the pair, this case has a total of 3 non-equivalent colorings.

- No pairs of opposite edges have the same color. Again, we will fix two thick edges and carefully count the possibilities.

These are the only possible choices, once we fix the thick edges, and they are all nonequivalent, since no rotation will keep the two thick edges in place. Furthermore, the first two tetrahedra each have a thick-thick-thin face, but the first one has a face with edges colored thick-thin-dashed (going clockwise), whereas the second has a thick-thin-dashed face (going counterclockwise). For the next two tetrahedra, both have thick-thick-dashed faces, but the first has a thick-dashed-thin clockwise face, but the last has a thick-dashed-thin counterclockwise face.

In sum, there are $2+3+4=9$ non-equivalent colorings.

Solution:

Brilliancy Prize Alternative Solution: Suraj Mathashery of Kennedy Middle School wrote a nice solution that was the only paper which correctly applied the "count all the colorings as a string and divide out by the symmetries" method. Many people correctly counted (exercise) that there are 90 ways to arrange the letters RRGGBB as a string, and then divided this number "because of rotations" to get an answer. The problem is that there is no single number one can choose! Here is a sketch of Suraj's method. You are urged to fill in the gaps.

- There are 90 colorings if we don't worry about any rotations.
- These fall into two categories: The colorings where all three pairs of opposite edges are colored the same way, and those colorings where this is not the case.
- The first category has just 6 colorings; the second has $90-6=84$.
- The first category has just three rotations that change it; while the second category has 12.
- Hence the number of non-equivalent colorings is $6 / 3+84 / 12=2+7=9$.