Suppose you have enough energy and resources to put together (in a momentarily static configuration in which they are all at rest at the same time) as many protons as you want to form a "proton star". This ball will last a split second before the protons repulsion disperses it back. But if made of enough protons, it will momentarily have enough mass to form a black hole (or am I wrong and there is no upper mass limit for this situation? So the question is, will a black hole form or not? My doubt is that it is always argued that a star collapses into a black hole because there is no other force in nature to counterbalance it, however notice that in this case, the repulsive force of of the charged protons is always larger than the implosive force due to gravity.
This question has already been answered by a Randall Monroe of XKCD and Dr. Cindy Keeler of the Niels Bohr Institute.
It forms a Naked Singularity. Which is an infinitely dense object, from which light can escape.
You Reissner–Nordström metric for this question, as opposed to the more well known Schwarzschild metric.
Overall, you will get a blackhole, but it would be a blackhole without an event horizon (a naked singularity). Naked Singularities are forbidden by General Relativity, which I assume is the model of framework we want to work within for this question.
I think this is one of those situations where it is cleanest to think relativistically and concentrate on gravity as spacetime geometry, not gravity as a force.
If the stress energy tensor's components reach a such a magnitude that an horizon forms, then the mass/energy inside the horizon is doomed to stay inside evermore. This is a question of spacetime curvature, the geometry becomes such that nothing can escape the horizon unless somehow it travels backwards in time. Forces, no matter how big, cannot change this result once the horizon forms; although they will change the trajectories of the sources in the stress energy tensor, yet you've posed your question in a way that the assumption that the protons are converging fast enough to reach this point is implicit. The critical speed / energy of convergence to achieve this assumption is finite.
You also make the point
the repulsive force of of the charged protons is always larger than the implosive force due to gravity.
Again, this is indeed true in Newtonian gravity where we think of gravity as a force. In Newtonian gravity space does not deviate from Euclidean geometry, so once we're tilting local light cones enough that they are contained within a horizon, then we're well beyond Newtonian validity. Again, you simply need to give the protons enough kinetic energy to come condensed enough to form a horizon, then you have one mightily charged black hole and whose steady state is described by the Reissner–Nordström metric.
The black hole apparently doesn't needfully end up charged. Apparently (this is well outside my knowledge), according to a comment by theoretical physicist Physics SE User Lewis Miller:
The scaling of the electromagnetic force is irrelevant because long before the protons reach the density suggested here the protons will beta decay into neutrons and positrons. The positrons, being much less massive will disperse and the remaining neutrons will form the black hole. This is not too much different from what happens in a supernova of a very large star.
As I said- outside my knowledge, but certainly seems sound and intuitively sensible to me. One of the things that I like about this comment and Lewis's answer is that they show we can clearly bring observations and knowledge of real astronomical objects to the discussion of the OP's question.
Also see the XKCD article cited in User Aron's answer.
If a Reissner–Nordström black hole has a large enough amount of positive charge then General Relativity foretells a naked singularity. Whether or not classical gravity can be trusted to this level is moot. However, note that this does not happen if the charge is assembled as in the OP's question; the large amount of energy needed to assemble the black hole in the first place contributes to mass energy $M$ of the black hole such that $r_s \gg 2\,r_q$ (in the notation of the Wikipedia Reissner-Nordström metric article); $r_s$ being the Schwarzschild radius and $r_q$ a radius term resulting from the Coulombic repulsion (this is the quantitative description of my qualitative statement that "once we're tilting local light cones enough that they are contained within a horizon, then we're well beyond Newtonian validity"). The condition $r_s \leq 2\,r_q$ is the condition for a naked singularity.
The strong force is responsible for neutron stars not becoming black holes. Although attractive at nuclear densities, it becomes repulsive at somewhat higher densities. This repulsion is many times greater than the electromagnetic repulsion between protons at similar densities. Yet we know that if a star is too large, it's supernova event will lead to a black hole rather than a neutron star. For star masses this large gravity crushes the strong force. Similarly, a proton star of this mass (if somehow assembled) will also be crushed into a black hole.