NEW ANSWER:
As there has been much confusion on this point (some of it mine...):
Definition: A Riemannian 2-manifold $S$ is of hyperbolic type if the universal cover of $S$ is conformally equivalent to the open unit disk, $D$.
On the other hand we have
Definition: A hyperbolic surface $S$ is a surface equipped with a complete Riemannian metric of constant curvature minus one.
It is an exercise to show that all hyperbolic surfaces are surfaces of hyperbolic type. On the other hand, a surface of hyperbolic type need not be hyperbolic. As an easy example of this, choose your favorite positive function $f$ on the disk $D$ and use $f$ to scale the Poincare metric. This new metric is (almost surely) not constant curvature but is conformally equivalent to the Poincare metric.
With these definitions in place: the original question is ill-posed. Knowing that a surface $S$ is of hyperbolic type does not suffice to tell us the metric. To be precise, there are conformally equivalent metrics $\rho_0$ and $\rho_1$ on the open disk $D$ so that the first is Gromov hyperbolic and the second is not. (Eg, let $\rho_0$ be the Poincare metric while $\rho_1$ has larger and larger "mushrooms" as you walk to infinity.)
OLD ANSWER (written in terms of the above definitions):
I'll assume that you are asking for a sufficient condition to ensure that a hyperbolic surface $S$ is Gromov hyperbolic. One condition is that $S$ has finite area. In this case $S$ has a compact core (which is of no interest in this setting) and a finite number of cusps. A cusp is obtained by modding out a horodisk by a parabolic isometry. All cusps are quasi-isometric to rays. Thus $S$ is quasi-isometric to a tree having one vertex and one ray per cusp.
A simpler condition is that $\pi_1(S)$ is finitely generated. The allowed surfaces are now somewhat more complicated: in addition to cusps there can be funnels in the complement of the compact core. A funnel is obtained by modding out a half-plane by a hyperbolic isometry. All funnels are quasi-isometric to the hyperbolic plane. (This is a nice exercise!) So, here, $S$ is quasi-isometric to the one-point union of a collection of hyperbolic planes and rays.
As for the "opposite direction": When the group is infinitely generated things can be very strange. For example, consider any cubic, connected graph $X$, of infinite diameter, where all edges have length one. (This is a very large class of metric spaces, even after passing to quasi-isometric equivalence classes.) Then, for any such graph $X$ there is a hyperbolic surface $S_X$ quasi-isometric to $X$.
The results on open Riemann surfaces are not "rare". They are just well forgotten.
I only list a few books which deal with open Riemann surfaces:
MR0114911, MR0228671, MR0159935, MR0264064, MR1973182.
It is true that there are "too many" Riemann surfaces, and not too much can be said about
"all of them". However, there is a highly non-trivial classification, and some subclasses
are very important. For example, the class of "hyperelliptic" surfaces of infinite
genus plays a very important role in the study of Schrodinger operators and their
finite difference analog, see for example the works of Sodin and Yuditskii, MR1288838,
and Kean, Moerbeke, MR0397076. Abelian covers of compact surfaces is another class
of open Riemann surfaces that was studied a lot: MR0740581.
Best Answer
As has been pointed out, the inequivalence of the three is elementary.
The original proofs of Koebe and Poincare were by means of harmonic functions, i.e. the Laplace equation ${\Delta}u = 0$. This approach was later considerably streamlined by means of Perron's method for constructing harmonic functions. Perron's method is very nice, as it is elementary (in complex analysis terms) and requires next to no topological assumptions. A modern proof of the full uniformization theorem along these lines may be found in the book "Conformal Invariants" by Ahlfors.
The second proof of Koebe uses holomorphic functions, i.e. the Cauchy-Riemann equations, and some topology.
There is a proof by Borel that uses the nonlinear PDE that expresses that the Gaussian curvature is constant. This ties in with the differential-geometric version of the Uniformization Theorem: Any surface (smooth, connected 2-manifold without boundary) carries a Riemannian metric with constant Gaussian curvature. (valid also for noncompact surfaces).
There is a proof by Bers using the Beltrami equation (another PDE).
For special cases the proof is easier. The case of a compact simply connected Riemann surface can be done by constructing a nonconstant meromorphic function by means of harmonic functions, and this is less involved than the full case. There is a short paper by Fisher, Hubbard and Wittner where the case of domains on the Riemann sphere is done by means of an idea of Koebe. (Subtle point here: Fisher et al consider non-simply connected domains on the Riemann sphere. The universal covering is a simply connected Riemann surface, but it is not obvious that it is biholomorphic to a domain on the Riemann sphere, so the Riemann Mapping Theorem does not apply).
The Uniformization Theorem lies a good deal deeper than the Riemann Mapping Theorem. The latter is the special case of the former where the Riemann surface is a simply connected domain on the Riemann sphere.
I decided to add a comment to clear up a misunderstanding. The theorem that a simply connected surface (say smooth, connected 2-manifold without boundary) is diffeomorphic to the plane (a.k.a. the disk, diffeomorphically) or the sphere, is a theorem in topology, and is not the Uniformization Theorem. The latter says that any simply connected Riemann surface is biholomorphic (or conformally equivalent; same in complex dimension $1$) to the disk, the complex plane or the Riemann sphere.
But the topology theorem is a corollary to the Uniformization Theorem. To see this, suppose $X$ is a simply connected (smooth etc.) surface. Step (1): Immerse it in $\mathbb{R}^3$ so as to miss the origin. Step (2): Put the Riemann sphere (with its complex structure!) in $\mathbb{R}^3$ in the form of the unit sphere. Step (3): For every tangent space $T_pX$ on $X$, carry the complex structure $J$ from the corresponding tangent space on the Riemann sphere by parallell transport (Gauss map) to $T_pX$. This is well-defined by choosing a basepoint and recalling that $X$ is simply connected. Step (4): Presto! $X$ is now a Riemann surface (it carries a complex structure), so it is biholomorphic to the disk or the plane or the Riemann sphere, thus diffeomorphic to one of the three.
Of course, I have glided over the question of immersing the surface in 3-space, because this is topology. Actually, I vaguely recall that there is a classification of noncompact topological surfaces by Johannsen (sp?), and no doubt the topological theorem would immediately fall out of that.