Yes, there are rules that depend on the quantum numbers carried by the particles under question and the energy available for the interaction.
In general we label as annihilation when particle meets antiparticle because all the characterising quantum numbers are equal and opposite in sign and add and become 0, allowing for the decay into two photons, two because you need momentum conservation.
A positron meeting a proton will be repulsed by the electromagnetic interaction, unless it has very high energy and can interact with the quarks inside the proton, according to the rules of the standard model interactions.
When a neutron meets an antiproton the only quantum number that is not equal and opposite is the charge, so we cannot have annihilation to just photons, but the constituent antiquarks of the antiproton will annihilate with some of the quarks in the neutron there will no longer be any baryons, just mesons and photons, and all these interactions are given by the rules and crossections of the standard model.
We agree that the pair has integral charge (zero in fact), right?
So now we need a model of hadronization (the process in which the quarks generate hadrons). You've probably got some picture of the quarks flying apart and then separately undergoing a process to produce a bunch of stuff.
Unfortunately that picture isn't really correct.
The principle of "confinement" is that colored objects (i.e. quarks) can not exist in isolation, so you should think of the pair of quarks being connected by a region of high strong-nuclear force potential (often called a "flux tube"). The nature of the strong force is that it get stronger as the separation between colored objects increases; so as the quarks pull further and further apart the the total energy in the flux tube grows at the pair separates until production occurs.
So you go from
$q$ -- $\bar{q}$
to
$q$ ----------- $\bar{q}$
to
$q$ ---- $x$ $\bar{x}$ ------- $\bar{q}$
to
$q$ ----- $x$ -- $\bar{x}$ ---- $\bar{q}$
and these may keep on separating from there (energy permitting, of course).
Depending on the nature of $x$ and $\bar{x}$ this could be the end of it, you get two mesons and you're done. But that is unlikely: more often one or both pairs are not color neutral or not bound so they pull apart some more and the new flux tubes also break.
For concreteness pretend that
- $q$ is red-up
- $x$ is antired-antidown
- the kinematics are such that $q$ and $x$ can end up bound, but $\bar{x}$ and $\bar{q}$ won't
this meas the next step might loop like
$u\bar{d}$ ------- $d$ ------- $\bar{y}$ -- $y$ ----- $\bar{u}$
and when the first pair binds
$\pi^+$ ------- $d$ ------- $\bar{y}$ -- $y$ ----- $\bar{u}$
But since we started neutral and each breaking produces a particle--anti-particle (which is neutral in total) the total will end up neutral. And it is confinement that insures that each particle is either a baryon or a meson (because those are the only color neutral bound states known (people are still hoping to see more complex states)).
So the picture that you should have is of extra pairs being formed in the middle until we run out of energy or everything has combined by lucky chance.
Best Answer
It depends on your definition of annihilation. But microscopically all processes are described by Feynman diagrams such as these
of which last one describes electron positron annihilation (if it weren't for the typo in the out-going photon). But as you can see it's all a simple matter of how you turn your head around and the very same diagram represents emission (or absorption) of photon by electron (or positron). Does electron annihilate with photon and create a brand new electron? You can certainly interpret it that way. In other words, it's just a question of terminology and interpretation. Actual physics doesn't depend on how you call the process. It is encoded in the underlying math of quantum field theory (QFT).
In any case, the punchline is that annihilation (in the strict sense of particle-antiparticle inelastic collision) doesn't have a special place in one's vocabulary once they learn their QFT and particle physics. It's just one particular kind of interaction. So you might as well ask which arbitrary interactions are allowed. And answer to that is: there's quite a lot of them and they are described by Standard Model. But the basic picture is that particles can be charged under certain charges: either the familiar electromagnetic, or less familiar weak and strong charges. Or in more modern language whether some families of particles form a multiplet under some gauge group. For weak force with group SU(2) you get lepton (e.g. electron-neutrino) and quark (e.g. up-down) doublets. For strong force with group SU(3) you need triplets and these are precisely the red green blue colors of quarks that you probably heard about.
In any case, for every multiplet there is a diagram like the ones above where you have two charged particles and one mediating particle between them (photon, weak bosons or gluons). Besides this you can also get funnier diagrams with e.g. three or four gluon lines. But that's it, these are all of allowed interactions of Standard Model.