Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology
The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question. 

The $a$ is the occurring in the Fourier transform is the scale factor which has a time dependence. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,

$$
\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q} 
$$

$$
 = \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$

$$
 = \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$

$$
 = \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$

The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel.