The answer in general is **no**, as shown by the following example.

Take the group $G=\mathbb{Z}/2 \mathbb{Z}$, acting on $\mathbb{A}^2$ as $$(x, \, y) \mapsto (-x, \, -y).$$ This action is free outside the unique fixed point $p=(0, \, 0)$, so $Y= \mathbb{A}^2-\{p\}$ is $G$-stable.

Furthermore, we have $$\mathbb{A}^2/G = \textrm{Spec} \, \mathbb{C}[x, \, y]^G=\textrm{Spec} \, \mathbb{C}[x^2, \, xy, \, y^2] = \textrm{Spec} \, \mathbb{C}[u, \, v, \, w]/(v^2-uw).$$
So $\mathbb{A}^2/G$ is an affine variety isomorphic to a quadric cone in $\mathbb{A}^3$, whereas $Y/G$ is this cone minus its vertex.