Yes. By Hilbert's theorem, the ring of $G$-invariant functions on $\mathbb C^n$ is finitely generated. Say the number of generators is $N$, then this gives a $G$-invariant holomorphic map to $\mathbb C^N$.
If one defines the quotient $\mathbb C^n/ G$ as the spectrum of this ring, then the map is clearly an embedding, and since, by computing the tangent spaces, the space being embedded is smooth, it is a smooth embedding. But if we take a complex-geometry perspective on what the quotient is, there is someting to check.
I claim this map is a smooth embedding. We will check this by checking that the map separates points and tangent vectors. If $P$ and $Q$ are two nonzero points of $\mathbb C^n$ that are not in the same $G$-orbit, then let $f$ be a function that vanishes at $P$ but not $Q$. The product of all the $G$-conjugates of $f$ is invariants, and separates $P$ and $Q$.
Similarly, if two tangent vectors at a nonzero point $P$ are distinct, we can find a function $f$ that vanishes at $P$ but not any $G$-conjugate point, and with a derivative that vanishes along one tangent vector but not the other. Here we use the fact that $G$ acts freely on $\mathbb C^n - 0$. Then multiplying all the $G$-conjugates of $f$ produces a function that vanishes on one vector but not the other. (Alternately, if the tangent vectors are in the same direction, we just need to find a map with nonvanishing derivative in that direction - the same construction works.