Yes, the theorem is:
Suppose that $G$ is a finitely-generated group whose Cayley graph quasi-isometrically (qi) embeds in $R^3$. Then $G$ is commensurable to a free abelian group of rank $\le 3$. (The converse is, of course, also true.)
In other words, $G$ contains a free abelian subgroup $A$ of finite index, so that $A$ has rank $\le 3$.
Here is a proof. First, you note that $R^3$ has polynomial growth, equivalent to $x^3$. Thus, $G$ also has growth at most $x^3$ since growth of a qi embedded subspace can only be lower than the one for the ambient space. By Gromov's polynomial growth theorem, it follows that $G$ is virtually nilpotent. For nilpotent groups there is a precise formula for growth in terms of their derived series, due to Bass and Guivarch. This formula implies that the group has to be virtually abelian of rank $\le 3$.
You can find definitions and proofs of most of the results in this book.
There is a bit more general result, due to Scott Pauls that if a nilpotent group $G$ qi embeds in, say, a Hilbert space, then $G$ is virtually abelian. However, in the 3d case you are interested in, you do not need this.