Hyperbolicity on Riemann Surfaces For Riemann surfaces there are at least to possible notions of hyperbolicity. The classical one given by the Uniformization Theorem, or equivalently the type problem, which essentially says that a simply connected Riemann surfaces is conformally equivalent to one of the following:


*

*Riemann Sphere $\mathbb{C}\cup\{\infty\}$ (elliptic type).

*Complex plane (parabolic type).

*Open unit disk (hyperbolic type).


On the other hand, given a Riemann surface one can asks if it is hyperbolic in the Gromov's sense. In other words, does there exists $\delta>0$ such that all the geodesic triangles in the surface are $\delta$-thin? 
It seems to me that this two notions of hyperbolicity are not equivalent and one can have counterexamples in both directions. For instance, the two dimensional torus $\mathbb{T}^2$ is hyperbolic in Gromov's sense (since it is compact), but it's also a quotient of the Euclidean plane by a free action of a discrete group of isometries and therefore, of parabolic type. 
My questions are: what is a sufficient condition for a surface of hyperbolic type to be Gromov's hyperbolic? what is known about the relation of these two notions?  
Related Question: Let $G$ be an infinite planar graph with uniformly bounded degree and assume that the simple random walk is transient. Is the graph necessarily Gromov's hyperbolic? 
 A: I don't quite understand your question about surfaces as the notions of hyperbolicity you are talking about deal with two different structures: a Riemann surface is endowed with a conformal structure, whereas Gromov's hyperbolicity is a property of metric spaces (in particular, of Riemannian manifolds). 
There is also some confusion in the torus example: when saying that it is of parabolic type you are talking about its universal cover, whereas when claiming that it is Gromov hyperbolic you are talking about the torus itself (actually it is not really fair to say that a compact metric space is Gromov hyperbolic).
Finally, concerning graphs the answer is no, because you can always attach to your favorite transient planar graph a sequence of circles with increasing radii - it won't change transience and planarity, but will prevent the resulting graph from being hyperbolic.
EDIT: I would still strongly advise against basing any examples or counterexamples on "compact hyperbolic spaces". Although formally they do have the $\delta$-hyperbolicity property, the whole point of developing this theory was to look at the large scale geometry of such spaces, and in particular at their behavior at infinity. If you wish, the notion of a compact hyperbolic space is as rich as that of a compact vector space. This is what I meant by saying that "it is not really fair to say that a compact metric space is Gromov hyperbolic". 
As for your revised question, one can formulate it in a more general way: when is a quotient of a Gromov hyperbolic space also Gromov hyperbolic? In order to see its scope, you may first look at the discrete case, where any regular graph is a quotient of the corresponding homogeneous tree. 
A: About the related question: it is a result of Babai that a (connected, locally finite) vertex-transitive, planar graph is isomorphic to the 1-skeleton of an Archimedean tiling of the sphere, or the Euclidean plane, or the hyperbolic plane. So assuming transience singles out the hyperbolic plane, and implies Gromov-hyperbolicity for the graph. See
http://www.cs.uchicago.edu/files/tr_authentic/TR-2001-04.ps
A: 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 $X$.
