This is a question about when a smooth complex algebraic space that is very close to being an algebraic variety is actually an algebraic variety.
General question: Let $X$ be a smooth separated complex algebraic space. Let $Z$ be a closed subspace of high codimension, such that the complement of $Z$ in $X$ is a quasi-affine algebraic variety. What additional conditions would suffice to conclude that $X$ is in fact an algebraic variety? In particular, I'd need to know why the regular functions on $X \setminus Z$, which should extend over $X$ by Hartog, suffice to separate points on $Z$.
Specific question: For me, $X$ will itself arise as the quotient by the free action (in the sense of group actions on schemes) of a connected semi-simple complex algebraic group $G$ on a smooth quasi-affine complex variety $Y$. Here then, $X = Y/G$ is a geometric quotient in the category of separated algebraic spaces. Furthermore I require that the restriction of the $G$-action to the complement of some high codimension closed $G$-stable subvariety in $Y$, call this open set $U \subset Y$, induces a geometric quotient in the category of varieties. Then $U \to U/G$ is a $G$-torsor with $U/G$ a smooth quasi-affine variety. Must the space $X = Y/G$, containing $U/G$ as large open subset with high codimension complement, then also be a quasi-affine algebraic variety? If not, is there a counter-example?