This is a good question. It turns out to be several different questions.
For example the hydrogen atom always has 1 (and not 1.2 or 1.5) electron quanta
This is not actually true. The hydrogen atom is by definition a charge eigenstate, but an energy eigenstate of hydrogen is only approximately an electron number eigenstate. The approximation is a good one because the atomic energy scale is small compared to the mass-energy of an electron-positron pair. For heavy enough atoms (hand-wavingly, those with $Z\gtrsim 1/\alpha$), you can "spark the vacuum," and the uncertainty in particle number is of order unity. My knowledge of QCD is not great, but I believe that, for example, a hydrogen atom does have an uncertainty in quark number that is of order unity.
Note that this in contrast to the photon field which is regularly found in photon coherent states which are superpositions of photon number states.
There is a number-phase uncertainty relation. Because of this, you can't get a classical field (one whose phase can be measured) made out of fermions. To get coherence, you need to have the density of particles (per unit volume $\lambda^3$) to be $\gg 1$, but this is impossible for fermions.
For massive neutral bosons, it's possible in principle to have a coherent wave, but because of the number-phase uncertainty relation, it will have to have a large uncertainty in particle number. 795
This means that it has a large uncertainty in energy.
But this kicks the can down the road. Why don’t we see, for example, a superposition state of multiple proton and electron numbers.
You can have this, for example, in beta decay. If you prepare the parent nucleus in its ground state, and then wait, the <PERSON> equation says that the wavefunction becomes a mixture of the decayed and undecayed states. (The decayed state includes the emitted beta and neutrino.) These states aren't uncommon in any definable objective sense, but normally when we do nuclear physics experiments, we measure the emitted beta as our way of detecting that something has happened. Then we get decoherence between the state of the detector that has seen a beta and the one that hasn't.
Or maybe out differently, why don’t we see stable bound atomic states that are superpositions of H and He atoms?
These systems have different baryon numbers. Baryon number is not only conserved, but we also don't have any observables $A$ such that $\langle \text{H} | A | \text{He} \rangle\ne0$. Therefore they inhabit different superselection sectors, and we can't make coherent superpositions of them. So there is nothing in quantum mechanics that prohibits the superposition, but the superposition has no observable consequences, e.g., you can't observe interference effects. 795