= Solution
Continue with $\eta=-g$ and dot products in that auxiliary <Minkowski metric>[metric]. Odd <Gamma-matrix trace>[gamma traces] vanish. The two current <tensors> are
$$
\begin{aligned}
H_{\mu\alpha}&=4[p_{2\mu}p_{1\alpha}+p_{2\alpha}p_{1\mu}-\eta_{\mu\alpha}(p_1\cdot p_2+M^2)],\\
L^{\mu\alpha}&=4[k_1^\mu k_2^\alpha+k_1^\alpha k_2^\mu-\eta^{\mu\alpha}(k_1\cdot k_2)].
\end{aligned}
$$
Contracting gives
$$
H_{\mu\alpha}L^{\mu\alpha}=32[(p_1\cdot k_1)(p_2\cdot k_2)+(p_1\cdot k_2)(p_2\cdot k_1)+M^2(k_1\cdot k_2)].
$$
Let $u=2M^2-s-t$. The <Mass shell>[on-shell products] are $p_1\cdot k_1=p_2\cdot k_2=(M^2-t)/2$, $p_1\cdot k_2=p_2\cdot k_1=(M^2-u)/2$, and $k_1\cdot k_2=s/2$. Substitution yields
$$
\boxed{\overline{|T|^2}=\frac{4g_s^4}{9s^2}[(M^2-t)^2+(M^2-u)^2+2M^2s]
=\frac{64\pi^2\alpha_s^2}{9s^2}[(M^2-t)^2+(s+t-M^2)^2+2M^2s].}
$$
The result is already averaged over incoming <Color charge>[colours] and <spins> and summed over outgoing ones. For $M\to0$, it becomes $4g_s^4(t^2+u^2)/(9s^2)$.
An additional check uses the <Center-of-momentum frame>[centre-of-mass] angle and <velocity> $\beta_t=\sqrt{1-4M^2/s}$:
$$
\overline{|T|^2}=\frac{2g_s^4}{9}\left[1+\cos^2\theta+\frac{4M^2}{s}\sin^2\theta\right].
$$
It tends to $4g_s^4/9$ at <Particle-production threshold>[threshold] and is positive throughout the physical region. Keeping both printed minus signs would instead give $4g_s^4[(M^2-t)^2+(M^2-u)^2-2M^2s]/(9s^2)$, displaying the incorrect <mass> contribution explicitly.
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