Does the gravitational strength of a black hole at its singularity change the more it pulls things into it?
In General Relativity the spacetime curvature isn't defined at the singularity, which is one of the reasons it's a singularity. You could say the spacetime curvature at the singularity is infinity, but outside of a mathematician's brain you can't use infinity in equations because it isn't a number. This is one of the reasons most physicists think that quantum gravity becomes important as you get very near to the singularity, and that quantum gravity will prevent the curvature becoming infinite. It would be nice if we knew exactly what this theory of quantum gravity looks like, but so far it's still a work in progress.
Even if you change your question to omit the words "at the singularity" and just ask:
this still isn't as simple a question as you might think. If you stay well away from the event horizon then if you add more matter to a black hole the spacetime curvature increases and it's gravitational field gets stronger, just as you would expect. The reason I'm being cautious is that observers who stay outside the event horizon will never see anything pass through the event horizon. If you stay well away from the event horizon this doesn't matter because Gauss' law ensures the gravity outside the black hole is the same whether the matter passes through the event horizon or not.
Re the comment: if you want to calculate something, like how fast an object falls into a black hole, you have to choose a set of co-ordinates to use. Even in Special Relativity things look different in different co-ordinates e.g. if we watch a fast moving spaceship we see the spaceships time running slow. But from inside the spaceship and using a different set of co-ordinates centred on the spaceship the spaceship's time runs normally. In General Relativity the difference between various co-ordinate systems gets even more extreme.
With black holes there are three common co-ordinate systems. Schwarzschild co-ordinates are what we outside the black hole think of as co-ordinates e.g. $t$ is the time we measure whith a clock and $r$ is the distance we measure with a ruler. The second set of co-ordinates are those used by someone falling freely into the black hole, and these are called freefall co-ordinates. Finally the third set of co-ordinates are those used by an observer hovering stationary near to the black hole. These are called shell co-ordinates.
Why am I ranting on about this? Well suppose you're sitting on one of the two objects falling into the black hole i.e. you're using freefall co-ordinates. You'll fall through the event horizon in a finite time and shortly afterwards meet a sticky end at the singularity. If you're falling freely you don't experience any gravity, though you will experince tidal forces. If you're on the second object then you will experience slightly greater tidal forces than on the first object because the first object's mass is added to the black hole. However unless the first object is extremely massive the extra tidal forces will be too small to measure. If you're sitting on the second object you will see the first object pass through the event horizon, but for a free falling observer there's no way to tell where the event horizon is other than by calculating where it should be. You don't feel any special forces as you pass the event horizon.
If you're well away using Scharzschild co-ordinates you'll see both objects slow down as they approach the event horizon, and unless you wait an infinite time you'll never see them fall through the horizon. You will see the second object experiencing a slightly greater gravitational field than the first because of the mass of the first object, but again the difference is likely to be too small to measure.