OK, I figured out the answer. Prahar is right that it it's due to a difference in the definition of the amplitude. But it was really not obvious to me why the amplitudes would differ by a kinematic factor, so I'll write out the full explanation in case it's ever useful to anyone else.
The amplitude that should be used in the second equation is not the normal transition amplitude, but the transition amplitude in the helicity formalism: $$T_{\lambda_a \lambda_b \lambda_c \lambda_d} = \langle \Omega' \lambda_c \lambda_d|T|\Omega\lambda_a \lambda_b \rangle.$$
I had assumed that we could just do a normal basis change from the helicity basis to a normal spin basis without picking up any weird kinematic factors, so that it was legitimate to treat $T_{\lambda_a \lambda_b \lambda_c \lambda_d}$ and $\mathcal{M_{m_a m_b m_c m_d}}$ in the same way. This is not the case, and it's because of the way the two-particle helicity states $|\Omega\lambda_a \lambda_b \rangle$ are defined. The details are given in the Spin Formalisms preprint by Chung. Eq 4.11 is the definition of the two-particle helicity state, Eq. 4.18 gives the normalization condition for these states, and Eq 4.19 gives the normalization constant $$a(p) = \frac{1}{4\pi}\sqrt{\frac{p}{w}},$$ where $p$ is momentum of one of the particles and $w=\sqrt{s}$ is the CM energy. Basically, with this convention, the standard normalization of the two-particle state differs from the normalization of (the outer product of) two one-particle states by a factor of $a$. So to get from the transition amplitude in the helicity formalism to the more standard normalization of the transition amplitude, we need to add a factor of $a(k)a(q)$, which makes the two equations above agree exactly.
