= Solution
Use the fundamental <SU(3)> <normalization> $\operatorname{tr}(T^aT^b)=\delta^{ab}/2$, with <Hermitian> generators. The average over the $3\times3$ incoming colours and sum over the outgoing colours gives
$$
\begin{aligned}
C_{\rm col}
&=\frac1{N_c^2}\sum_{i,j,k,l}\left|\sum_a(T^a)_{ji}(T^a)_{kl}\right|^2\\
&=\frac1{N_c^2}\sum_{a,b}\operatorname{tr}(T^aT^b)\operatorname{tr}(T^aT^b)
=\frac{N_c^2-1}{4N_c^2}=\frac29\qquad(N_c=3).
\end{aligned}
$$
The complete <spin> and colour averaged result is consequently
$$
\boxed{\overline{|\mathcal M|^2}
=\frac{4g_s^4}{9}\frac{t^2+u^2}{s^2}
=\frac{64\pi^2\alpha_s^2}{9}\frac{t^2+u^2}{s^2},\qquad
\alpha_s=\frac{g_s^2}{4\pi}.}
$$
For massless two-body centre-of-momentum <kinematics>, $t=-s(1-\cos\theta)/2$ and $u=-s(1+\cos\theta)/2$, so the same answer is $2g_s^4(1+\cos^2\theta)/9$. This checks positivity, the symmetry under exchanging the outgoing directions, and the absence of a forward $t$-channel singularity in this particular process.
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