Let $X$ be an affine $G$-variety where $G$ is a reductive group. All the varieties are over $k$ , where $k$ is a field (if it necessary we can assume it is algebraically closed). It is a known theorem that there exists an affine good quotient $p:X \to X//G$. Let us call $X//G=Y$.
I'm interested in the following situation:$Z \subseteq Y$ is a closed subset (considered as a subvariety with the standard reduced structure). We consider the closed $G$ invariant subset $p^{-1}(Z)=W$ of $X$: I'd like to prove that the map $p:W \to Z$ is actually a (categorical) quotient.
This accounts to prove the following:we call $B$ the $k$ algebra $\mathcal{O}_X(X)$. We then have $\mathcal{O}_Y(Y)=B^G$. With this identification, the subscheme $Z$ corresponds to an ideal $I \subseteq B^G$ while $W$ corresponds to $\sqrt{IB}$. We then want to prove that $$\dfrac{B^G}{I}=\left(\dfrac{B}{\sqrt{IB}}\right)^G .$$
We can think of $\sqrt{IB}$ as the function $f:X \to \mathbb{A}^1_k$ such that $f(p^{-1}(Z))=0$. It should then be true that $\sqrt{IB} \cap B^{G}=I$:on the left hand side we have the $G$ invariant functions $f:X \to \mathbb{A}^1_k$ such that $f(W)=0$; these functions factorize through $Y$ to a function $f \in B^G$ such that $f(Z)=0$ so that $f \in I$.
Now, I would like to follow the approach used by Dolgachev in its Introduction to geometric invariant theory on page $44$ : he uses that reductive groups are actually geometrically reductive i.e given a representation $V$ and a fixed non zero vector $v \in V$, there exists a homogenous $G$ invariant polynomial $F$ on $V$ sucht hat $F(v) \neq 0$. More precisely, if char $k=0$ we can assume $F$ to be linear and Dolgachev also proves the equality of invariant rings wanted.
If char $k=p$, it is actually true that we can assume $F$ to be homogenous of degree $p^r$ and that for every $a \in \left(\dfrac{B}{\sqrt{IB}}\right)^G$ there exists an $h_a \in A^G$ such that $h_a-a^{p^r} \in I$. This does not really imply the claim, however it seems very close to me. I do not know how to conclude however.