We can answer your question by considering the first law of thermodynamics on a body with uniform temperature $T$, which is only losing thermal energy via radiation:
$$ \frac{dU}{dt}=\dot{Q} - \dot{W}$$
where
$$ U = \rho c_p T = \frac{m}{V}c_pT$$
$$ \dot {Q} = -\epsilon \sigma T ^4$$
$$ \dot{W} =0 $$
where $ \dot {Q} = -\epsilon \sigma T ^4$ is the emitted radiation which depends on the emissivity $\epsilon$.
Substituting into the first law,
$$ \frac{m}{V}c_p \frac{dT}{dt} = -\epsilon \sigma T ^4$$
Integrate from $t=0$ to $t$, noting that $T(0) = T_0$, and solve for $t$:
$$ t = \frac{3 m c_p}{V \epsilon \sigma}\left(T^{-1/3} - T_0^{-1/3}\right)$$
You can see that heat capacity definitely affects how fast the body cools. Specifically, larger heat capacity means longer cooling times since more energy is stored in the body at a given temperature.