show/hide this revision's text 5 hyperbolic versus hyperbolic type

EDITED

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):
show/hide this revision's text 4 discussed the converse

EDITED:

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 $X$.
show/hide this revision's text 3 added 6 characters in body

EDITED:

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 isometric 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.

When the group is infinitely generated things can be very strange.
show/hide this revision's text 2 Made correction as indicated by Agol.
show/hide this revision's text 1