fate of a hadron in a big rip As I understand it, if you try to "pull apart" a hadron with sufficient force, you just make more hadrons. Would the same thing happen in a Big Rip? (this answer suggests it would, as long as the expansion is sufficiently violent)
Suppose you start with a neutron (it doesn't matter what else is nearby, it will end up very far away) and a Big Rip happens. Many particles are produced from the original neutron. Then for some reason the expansion of space slows down to a reasonable level (call it "inflation" instead of "big rip"). After a short time most of the unstable particles will decay. What will be left? Will it be mostly protons, electrons and antineutrinos? Or an equal mixture of "matter" and "antimatter"?
 A: Since you hypothesise the big rip effect on the neutron, I am going to follow your scenario without questioning it and only address your questions about the outcome. In any case, I do not think that there is a consensus about the effect of the big rip at the scale of a neutron: a quantum theory of gravity would be needed to make sensible predictions, and we don't have one, as everybody knows. Finally, it goes without saying that the following is a very (very) naive description that can only convey a qualitative big picture.
The strong interaction has the property of confinement, which in the case of trying to separate a pair of quark $q\bar{q}$ is adequately pictured by a colour tube stretching between the quarks until it breaks, producing new $q'\bar{q}'$ pairs. I have only represented one breaking point but there can be many. It should be noted that this is not just a pretty pictures: high energy physicists use such a phenomenological model to simulate how the quarks produced by high-energy collisions at accelerators such as Tevatron, LEP, LHC, etc do eventually hadronise (i.e. produce hadrons in the final state) [*]. 

For the three valence quarks of the neutron, the colour tube configuration will look more like the following. I have reused the same letter q for all quark pair produced by the breaking of the colour strings so as to keep the diagram readable but they can of course be different. This time, I have shown multiple breaking points. I have shown this diagram to motivate that the ripping should produce baryons, not anti-baryons, and also mesons and anti-mesons, roughly in equal numbers since the quark-antiquark pairs can be made of any flavour. As the big rip continues, the mesons and anti-mesons would themselves split into more mesons as in the previous paragraph.

The final decay products will be stable particles, those you cited, but also a few others. Assuming there is no physics beyond the Standard Model: protons, antiprotons, electrons, positrons, (anti)neutrinos (any flavour) and photons (mostly from the decays of $\pi^0$). 


*

*The mesons will predominantly decay into protons, electrons, (anti)neutrinos and photons whereas the anti-mesons will predominantly decay into anti-protons, positrons, (anti)neutrinos and photons. And the neutral ones into an electrically neutral combinations of those. Thus with the same starting number of meson and anti-meson, this will lead to an equal mixture of matter and antimatter. 

*The baryons produced by the big rip will predominantly decay into protons, electrons, photons and (anti)neutrinos.


Thus whether the final ratio matter/antimatter is close to 1/2 really depends on the ratio of mesons to hadrons initially produced by the big rip. As explained pictorially above, the latter is greater than 1 (to give a more quantitative example, at LHC, the ratio of $p+\bar{p}$ over $\pi^++\pi^-$ ranges from less than 0.05 to about 0.3). So I would say that we should end up with some more matter than antimatter actually.  
[*] They are called QCD string fragmentation models and they are the foundation of Monte Carlo event generators such as PYTHIA, which can simulate all possible collisions.
A: Answering what I believe to be your implicit question:
There has been considerable speculation (particularly in the lay community) that at the point where the Big Rip / Dark Energy becomes relevant at the scale of mesons and hadrons, it may effectively cause every bound quark in the universe to become the center of a new Big Bang (i.e. a vast cascade of rapidly expanding particles and energy). The idea is charming: every atom in your body would give birth (22 billion years from now) to a myriad of new universes....  
I have spoken with some colleagues on this point and while they agree that it is an intriguing notion, they also agree it is impossible to seriously check on without a quantum theory of gravity. 
Even at an easier level of just considering possible solution of (for example) the Schrodinger Equation, the situation is far from simple. We have, effectively, two separate (but interacting) potential energy functions: the gravitational potential energy (largely repulsion caused by Dark Energy) and the Strong Force / QCD (attractive) potential between the quarks. Every time a new quark is 'created' the potential energy situation changes (some of the potential energy becomes mass; there is more mass, therefore more 'positive' gravity; there is a new particle to interact with the other particles via the Strong Force...) Trying to find a time-evolving solution of the equations with such an extreme (and dynamic) potential function would be far from easy.
In short, we don't really know what would happen. It is worth speculating (and is nice to think) that all the mass that would result would somehow overpower the repulsive effects of the Dark Energy and the universe (or at least that section of it) could settle down into a relatively stable state of expansion once again (and probably stay that way for about 36 billion years, until it has its own Big Rip, and so on); but we lack both a theory of gravity and probably adequate numerical tools to really analyze the situation at this time.   Hopefully we'll have it figured out sometime within the next 22 billion years. :)
