No, they don't. The uncertainty principle places limits on measurements of non-commuting observables on the same system. We are not talking about that here. See below.

> You can't know the complete state going in to the system

Sure you can: just prepare the system in a known state. For example, pass your photon through a vertically oriented polarizing filter: if it comes through, it must be vertically polarized, so you have complete knowledge of its polarization state. (You might have to try multiple photons to get one that passes through: that's why photon sources in these experiments are often inefficient.)

> or the complete state going out

Sure you can: you measure it. For example, you pass the vertically polarized photon that just came through your vertical polarization filter through a beam splitter, and you have detectors at each output of the beam splitter. Exactly one detector will fire for each photon.

> If the emitter were to only emit useable photons when it's "in the right state", what stops the "right state" for emitting photons to become correlated with the polarizers?

> There are a bunch of "unusable" photons bouncing around interacting with everything and transporting global state.

It looks like you don't have a good understanding of how the "emitter" works. What you are calling the "emitter" is really a filter, like the vertical polarizer described above: it throws away the photons coming from a source (like a laser) that don't meet a particular requirement (like vertical polarization). The thrown away photons are either absorbed (as in the case of the polarizer) or they just pass through the apparatus altogether and fly away (as in the case of parametric down conversion, for example: only a small percentage of the laser photons will be down converted, the rest just fly away and are gone).

In no case are the photons not used kept "bouncing around". They're gone. And the photons in the "right" state are just the ones that make it through the filter and are therefore in a known state when they come out, because that's how the filter works: the filter is uncorrelated with what's inside the experiment because, again, that's how the filter works (and it is tested to make sure it works that way).

> What happens to the photons that reflect off of the polarizers and travel back into the emitter?

There aren't any. See above.

> If a photon bounces off of a mirror it had to have 1. transfered momentum to whatever it hit, and 2. induced a sufficiently strong opposing electromagnetic field to cause the photon to be reflected or re-emitted.

1. Yes, but in these experiments the mirror is fixed to the Earth, so the momentum is transferred to the Earth, which means it's effectively gone. The entire Earth is not going to have a "memory" that can become correlated with the rest of the experiment.

2. No. You are thinking of it classically, but we are not talking about a classical process.