Yes, of course I knew that. I can't believe that I said something this moronic. I had an idea in mind which goes beyond the stupidity of what I wrote; if I can make it work I'll post it, hoping not to embarrass myself again :-)

If $I$ is the kernel of the map $R^h \otimes_R R^h \rightarrow R^h$, then $I/I^2$ is the module of differentials of $R^h/R$, which is $0$. Hence if $I$ is not $0$, it can not be finitely generated. But it seems clear that $(x + 1)^{1/2} \otimes 1 - 1 \otimes (x + 1)^{1/2}$ is not $0$.

If the gerbe is étale, then the fact that the cohomological dimension of the Galois group is 1 does imply that it is neutralizable (see for example the proof of Lemma 4.4) in my paper with Giulio Bresciani "Fields of moduli and the arithmetic of tame quotient singularities". In the general case I am not sure how to proceed.

This is a nice argument!

In case the gerbe has affine diagonal this is Theorem 8.1 in Di Proietto, Tonini and Zhang, Frobenius fixed objects of moduli. Actually, they show this for general fpqc gerbes, without finiteness hypotheses.

This is false in many cases, even with extremely mild singularities; for example, the fundamental group of the cone over a smooth conic in P^2 with the vertex removed is cyclic of order 2, while a resolution of singularities is simply connected.

$X$ is homotopy equivalent to $X_0$. See for example mathoverflow.net/questions/264940/…