How to see the various hyperbolic $n$-spaces $\mathbb{H}^n$ aren't quasi-isometric?
A helpful invariant for $\delta$-hyperbolic spaces is Gromov boundary: that is quasi-isometry of hyperbolic spaces induce homeomorphisms on the boundary, so if they have different homeomorphism types of boundary then they are not quasi-isometric. The boundary can be thought of as equivalence classes of rays, where rays are equivalent if they eventually stay within bounded distance from each other. From that definition it is clear that boundary of $\delta$-hyperbolic spaces agrees with the standard boundary you would put on Hyperbolic $n$-space.
Hyperbolic $n$-space (defined for $n>1$) has $S^{n-1}$ as the Gromov boundary, and you can tell spheres apart by a homology argument if you don't trust different dimensional spheres are not homeomorphic. So they are not quasi-isometric to each other.
Is not to hard to see that finitely generated free groups, on 2 or more generators, will have a Cantor set as it's boundary, which is not homeomorphic to spheres (it isn't even connected), so free groups are not quasi-isometric to any hyperbolic $n$-spaces.
As discussed in the comments, you could think of the real line as 1-dimentional hyperbolic space, in which case the free cyclic group is quasi isometric to 1-dimentional hyperbolic space. Note the free cyclic group you have 2 points in the boundary, which is a zero sphere and is note homeomorphic to the cantor set or higher spheres, so it is not quasi-isometric to other free groups.
There is a notion of "asymptotic dimension", defined for all metric spaces, which for $\mathbb{H}^n$ (or the Euclidean $n$-space) equals $n$ ($*$). It is monotone under coarse embeddings. In particular, $\mathbb{H}^n$ does not coarsely embed into $\mathbb{H}^m$ if $n>m$.
This latter fact does not follow from the Gromov boundary, because the latter is not functorial for coarse embedding (from the latter one gets a weaker statement, since anyway it's functorial for QI-embeddings: $\mathbb{H}^n$ does quasi-isometrically embed into $\mathbb{H}^m$ for $n>m$ (and in particular they are not quasi-isometric, as mentioned by Paul).
Also, $\mathbb{H}^n$, $n\ge 2$ does not coarsely embed into the free group: this follows from asymptotic dimension (which is 1 for the free group). An easier way to see that they are not quasi-isometric consists in using that the space of ends is a Cantor space in the first case an a point for $\mathbb{H}^n$. Since the space of ends is functorial for coarse embeddings, this probably can help show that $\mathbb{H}^n$ does not coarsely embed into a free group but it doesn't seem to follow from a purely formal argument.
The boundary can be useful beyond its topology. For instance, the negatively curved symmetric spaces $\mathbb{H}^{2n}$ and $\mathbb{H}^n(\mathbf{C})$ are not quasi-isometric, although they have the same number of ends (1), the same asymptotic dimension ($2n$) and homeomorphic Gromov boundaries (a $(2n-1$)-sphere). The point is that their Gromov boundaries are not quasi-symmetrically equivalent.
($*$) I can only provide references, it's not obvious. In Chapter 10 of Buyalo-Schroeder's book, it's proved that the asymptotic dimension of a $n$-dimensional Hadamard manifold is $\ge n$. This applies to both the Euclidean and the real hyperbolic space.
In the same book they only provide non-sharp upper bounds. But Higes-Peng (arXiv link), Assouad-Nagata dimension of connected Lie groups. Math. Z. 273 (2013), no. 1-2, 283–302. proved that the asymptotic Assouad-Nagata (aAN) dimension of a connected Lie group $G$ with maximal compact subgroup is $\dim(G/K)$, this applies in particular to the connected part of the isometry group of any $n$-dimensional isometry-homogeneous contractible Riemannian manifold to show that its aAN dimension is $\le n$. It it trivial from the definition that the aAN dimension is an upper bound for the asymptotic dimension.
Combining, the asymptotic dimension (and the aAN dimension) of any homogeneous Hadamard manifold is $n$.