I believe this is where your thought experiment goes wrong:
However, the nuclei are so small and so widely separated that presumably they just orbit each other as the electron clouds annihilate
Why would they be widely separated? Solids touch in order of nanometers. The protons and antiprotons would attract each other with the 1/r^2 electric field, which is very strong. Even if the energy of the projectile is small enough to form "protoniums", ie. have antiprotons in the place of electrons in the quantum mechanical solution, the wave function will be so large that the protons antiproton wave function will overlap and annihilate. The lifetime is of the order of .1 to 10 microseconds.
To get an idea of the annihilation Here is a bubble chamber photo of an antiproton entering from below annihilating on a proton at rest int he chamber. The multiplicity is on pions and kaons which will then decay with their characteristic decays.
The multiplicity is high even for annihilation at rest.
If the projectile has high enough energy, it will be a scattering with the cross section of antiproton on proton, depending on the area, to get the percentage annihilated. It is very difficult to get a plasma of protons and antiprotons anyway. Annihilations have a very large crossection. The antiprotons will either scatter elastically if they have enough energy, and leave the samples or annihilate if in the region of protons.
If the two bits of matter are in space with only their own gravity, then some will escape a first scatter and then fall back and annihilate or escape if they have escape velocity.
Quantum field theory does not offer a description of "how" its processes work, just like Newtonian mechanics doesn't offer an explanation of "how" forces impart acceleration or general relativity an explanation of "how" the spacetime metric obeys the Einstein equations.
The predictions of quantum field theory, and quantum electrodynamics (QED) in particular, are well-tested. Given two photons of sufficient energy to yield at least the rest mass of an electron-positron pair, one finds that QED predicts a non-zero amplitude for the process $\gamma\gamma \to e^+ e^-$ to happen. That is all the theory tells us. No "fluctuation", no "virtual particles", nothing. Just a cold, hard, quantitative prediction of how likely such an event is.
All other things - for instance the laughable description in the Wikipedia article you quote - are stories, in this case a human-readable interpretation of the Feynman diagrams used to compute the probability of the event, but should not be taken as the actual statement the quantitative theory makes.
There is no "how", what happens between the input and the output of a quantum field theoretic process is a black box called "time evolution" that has no direct, human-readable interpretation. If we resolve it perturbatively with Feynman diagrams, people like to tell stories of virtual particles, but no one forces us to do that - one may organize the series in another way, may be even forced to do so (e.g. at strong coupling), or one may not use a series at all to compute the probability. The only non-approximative answer to "how" the scattering processes happen in quantum field theory that QFT has to offer is to sit down and derive the LSZ formula for scattering amplitudes from scratch, as it is done in most QFT books. Which, as you may already see from the Wikipedia article, is not what passes as a good story in most circles.
But neither nature nor our models of it are required to yield good stories. Our models are required to yield accurate predictions, and that is what quantum field theory does.
Best Answer
A particle isn't really a point particle; its position is best described by a wavefunction: the probability of finding it in any particular region in space.
For annihilation to occur the wavefunctions of the two particles have to overlap - and to the extent that they overlap there will be a probability that annihilation can occur. The greater the overlap, the greater the probability. "Overlap", in this context, is the integral of the product of the wavefunctions over all space.
This is the point of the answer to the question you linked; now you are asking (in essence) "what is the extent of the wave function"?
This is of course a function of the mass of the object - the uncertainty principle tells us that $\Delta x \Delta p > \hbar$. The better known $p$, the greater the uncertainty in $x$. Or - the lighter the particle, the larger the uncertainty in its position.
I'm not sure what uncharged particle/antiparticle pair you are thinking of...