There is no reason to expect positive answer to Q1, even for a linear action with a dense orbit. Let $X$ be the space of $n\times m$ matrices over a field $K$ and $G={\rm GL}_n(K)$ acting by left multiplication. Then by basic linear algebra $A$ and $B$ are in the same orbit iff $\ker A=\ker B$. For $n>m$, the matrices with zero kernel (i.e., of rank $m$) form a single dense orbit. However, the orbits of $G$ on $X$ are parametrized by pairs $(r, L)$, where $L$ is an $r$-dimensional subspace of $K^m$. In particular, the orbit space is infinite.

There is no reason to expect positive answer to Q1, even for a linear action with a dense orbit. 

Let $X$ be the space of $n\times m$ matrices over a field $K$ with $G={\rm GL}_n(K)$ acting by left multiplication. By basic linear algebra $A$ and $B$ are in the same orbit iff $\ker A=\ker B$. Now suppose that $n\geq m\geq 2$. Then the matrices with zero kernel (i.e., of maximal rank, $m$) form a single dense orbit in $X$. However, the orbits of $G$ on $X$ are parametrized by pairs $(r, L)$, where $r\leq m$ and $L$ is an $r$-dimensional subspace of $K^m$. In particular, the orbit space is infinite.