I am just recording what was said in the comments so this question does not appear *completely* unanswered.

Let $\pi_X:X\to X//G$ be the GIT quotient of an affine variety over $\mathbb{C}$ by a reductive group $G$. WLOG assume the action is effective.

First, a point is properly stable if its orbit is closed and it has finite stabilizer. The locus of properly stable points is Zariski open. Then since each fibre $\pi_X^{-1}(\pi_X(x))$ is a union of orbits, this union contains a unique closed orbit, two such orbits are in the same fibre if and only if their closures intersect, and such intersections can be detected by 1-parameter subgroups, we can conclude that if $x$ is properly stable then $\pi_X^{-1}(\pi_X(x))$ is set-theoretically the orbit $Gx$.

Please see *Stability of Affine G-varieties and Irreducibility in Reductive Groups* by Casimiro and Florentino as a reference.

As pointed out by Jason Starr in the comments, the fibre is not generally scheme-theoretically the orbit however. A counter-example is the action of $\mathbb{G}_m$ on $\mathbb{A}^2\times \mathbb{G}_m$ by $s\cdot((x,y),t)=(sx,sy,s^{-n}t)$ for $n>1.$ As noted by the OP, this is apparent even for $n=2$.

We now refer to the Luna Slice Theorem; see Luna’s slice theorem and applications by Drézet. Let $V$ be a slice at a properly stable point $x$, and let $\pi_V:V\to V//S$ be the corresponding quotient where $S$ is the stabilizer of $x$ (necessarily a reductive subgroup). Then there is an isomorphism: $$G\times_S \pi^{-1}_V(\pi_V(x))\cong \pi_X^{-1}(\pi_X(x)).$$

So, the fibre is the scheme-theoretic orbit if it is smooth which, by the Chevalley-Shephard-Todd Theorem, occurs if and only if the stabilizer is generated by pseudoreflections.