Consider first the set $S_m$ of all possible clock faces using $m$ colours, where we consider two to be equal only if the colours match without any rotation: there are $m^{12}$ elements in $S_m$. Let $f$ be the map that rotates an element of $S_m$ by $1/12$ of a complete circle, i.e. 30 degrees. ($f$ depends on $m$, but the action is the same regardless of $m$, so we just write $f$.) All elements $x \in S_m$ have period 12 under $f$, but some have smaller periods. Let $p(x)$ be the least period of $x$, and let $n_m(k)$ be the number of elements $x \in S_m$ such that $p(x) = k$.

It is clear that the possible values of $p(x)$ are the divisors of 12, and that the number of distinct clock faces, allowing at most $m$ colours and counting rotated clock faces as being identical, is
$$N(m) := \frac{n_m(12)}{12} + \frac{n_m(6)}{6} + \frac{n_m(4)}{4} + \frac{n_m(3)}{3} + \frac{n_m(2)}{2} + \frac{n_m(1)}{1}.$$

Since there are three ways to select two of the three colours, the answer to the problem is $N(3) - 3N(2) + 3N(1)$; the last term is because each one-colour clock face is removed twice, so we must add it back.

To make the solution explicit, it remains to compute a formula for $n_m(k)$ for each divisor $k$ of 12. It is clear that $n_m(1) = m$.

Next, $n_m(2) = m^2 - m$: we remove from the set of elements with period 2 those that also have period 1. Similarly, $n_m(3) = m^3 - m$ and $n_m(4) = m^4 - m^2$.

With a little more thought, we see that $n_m(6) = m^6 - m^3 - m^2 + m$ and $n_m(12) = m^{12} - m^6 - m^4 + m^2$. For instance, for $n_m(12)$, we remove from $S_m$ those elements with period 6 and those with period 4, leaving only elements with least period 12. However, when we removed the elements with smaller period, we double-counted those with least period dividing $\gcd(6, 4) = 2$, so we add these back.

Explicitly, we obtain $N(1) = 1$, $N(2) = 352$, $N(3) = 44,368$ and so the final answer is $N(3) - 3N(2) + 3N(1) = 43,315$.