@DavidLampert I'm not sure that your $R$ from above is perfect, is it? To me it seems as though any perfect $R$ is seminormal by the following argument. Suppose that $x$ is in $Q(R)$ is such that $x^3$ and $x^2$ are in $R$. Then, writing $p=2r+3$ we have that $x^p=x^3\cdot (x^2)^r$ belongs to $R$. Since $R$ is perfect there exists some $y$ in $R$ such that $x^p=y^p$. But, then $(x-y)^p=0$ which as $Q(R)$ is reduced implies that $x=y$, so $x\in R$. So, $R$ is seminormal.

@WilberdvanderKallen Hey Wilberd, this does seem to be a useful collection of results! Can you clarify how it helps precisely though? It seems as though what you call 'power surjective' is what Alper calls 'adequate' and so one can use results in your paper in lieu of Alpers to argue as I did in the main body of the question. Is that what you meant, or did you have something else in mind?

@PeterScholze Hmm, OK. Yeah, you might be right -- I perhaps forgot what Mumford assumed when he talked about these things. In any case, many thanks. I'll think carefully about what you suggested.

@PeterScholze Also, does one need Haboush's theroem to say that the closed points on on the affine GIT quotient correspond to closed orbits on the original? Doesn't that follow from general GIT nonsense? Even then I need to think of why the comparison is clear -- we're claiming that for a colimit of algebras the closed orbits of the colimits are the colimit of the closed orbits. This is probably simple, and I will think on it. Thanks again!

@PeterScholze Hey Peter, thanks for your quick response! On the generic fiber I agree things are simple. For the claim about universal homeomorphism, I am still slightly confused, but am probably just being thick. They are finite type $\mathbb{Z}_\ell$-algebras (from the strengthening of Mumford's theorem alluded to above) so for universal bijectivity it's reasonable that we need only check on $\overline{\mathbb{F}}_\ell$-points. But, I don't see how we're getting the homeomorphism part. Above it was suggested this map is obviously finite --this is the key basic thing I am missing.

@DavidBenjaminLim Hey man. Thanks for taking the time to read my post! I agree that it seems, in any case, that Alper's paper (or maybe at least Seshadri?) is needed here. Also, thanks for the typo -- I have fixed the lemma number.

Is there a well-studied notion of 'holomorphic Brauer groups'?

This is probably a question better suited for MO. Any questions involving removal of Noetherian hypotheses are pretty niche, and are much more likely to receive traction at MO.

@LSpice Yeah, I spent hours trawling though Denef, Loeser, Clucker, etc. I know it's buried in there morally but I can't find an exact statement. I need it in a more general context, but certainly Harish-Chandra et. al need it for the full-measure of G^reg(F), so I imagine they must also write it down somehwere. Thanks again!

@LSpice Hey Loren. It's because he's starting with a smooth model over $\mathcal{O}_K$ and that $Z$ also has a smooth model over $\mathcal{O}_K$. If you look at his proof I don't really think this gets used, and it does show my desired claim. But, because he technically starts with those assumptions it's a pretty non-canonical reference for a beginner. I was hopting there was just some fairly basic reference to this result (e.g. it's sketched in a paper of Harish-Chandra) or that someone could give a comprehensive proof here that I could reference/copy. This is for a set of notes I'm writing.