Ivella asked: "How do you calculate if the object will ever receive the light signal before reaching the center of the BH?"
The photon doesn't always reach the falling object. In the classic Schwarzschild coordinates both's $\rm r$ converge to $\rm r\to 2$ (in natural units) at $\rm t \to \infty$, but they are always at least an infinitesimal $\rm \Delta r$ apart from each other if the initial distance was too large, and the $\rm \Delta t$ is undefined since $\rm \infty \pm \Delta t=\infty$. You can better see it in Raindrop or Finkelstein coordinates.
Here the photons are pink while the other colors are for free falling particles with different initial velocities; for example, the yellow particle at $\rm r=2$ never gets overtaken by the last photon entering the frame at $\rm r=3$:

Ivella wrote: "I have the gut felling that the object will always receive it before reaching the event horizon, but I would like to know what the math says."
In Finkelstein coordinates the coordinate velocity for an ingoing photon is $\rm dr/d\bar{t}=-1$ and for a free faller from infinity $\rm dr/d\bar{t}=\sqrt{2} \ (r-2)/(\sqrt{8}-r^{3/2})$.
In Raindrop coordinates we have $\rm dr/d\acute{t}=-1-\sqrt{2/r} \ $ for the photon and $\rm dr/d\acute{t}=-\sqrt{2/r} \ $ for the free faller.
If you integrate that to get the $\rm r(t)$ or $\rm t(r)$ you see some photons never catch up with some free fallers (see here for photons, here for free fallers and here for an overview, related topics are here and here).
If we don't restrict ourselves to Schwarzschild black holes but also include rotating or charged black holes with an internal Cauchy horizon it is possible for the particle to end up on a trajectory on which he gets fried by all the photons entering after him at his position with infinite blueshift though, but that's a different story.