4 added 3 characters in body

Yes, this is true. Topologically, one may find a locally finite collection of properly embedded arcs in a connected surface whose complement is homeomorphic to $R^2$. Then make each of these arcs geodesic in the hyperbolic metric. The complement will be the fundamental domain of the type you want.

Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a classification of connected surfaces. In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface $\Sigma\cong S^2-X$ is obtained by removing a totally disconnected compact set $X\subset S^2$ from $S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set $X\subset S^2$ to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints) $X\subset I\subset S^2$. Then the properly embedded arcs $I\cap (S^2-X)$ give a decomposition of $\Sigma$ into $R^2$.

If the surface is non-planar, then one removes from $S^2-X$ a properly embedded countable collection of disks $D_1,D_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on $I$, and that the identifications either identifies antipodal points, or identifies two disks which are adjacent along a component of $I-X$. I-X$with a$\pi$twist. The complement$U=S^2-(I\cup_i D_i)$is again homemorphic to$R^2$. If we identify antipodal points of$D_j$, then this identifies two arcs in the boundary of$U$to obtain an open Mobius strip. We add two arcs connecting antipodal points of$D_j$to a point$x\in X$at the end of the interval of$I-X$which intersects$D_j$, which forms a single arc after identification of antipodal points of$D_j$, and cuts the Mobius strip back up into$R^2$. If adjacent disks$D_i, D_{i+1}$are identified, then the complement of the arcs$U$gives a thrice-punctured spherepunctured torus. We add 4 arcs connecting these points to$x$(again,$x\in X$is at the end of the interval of$I-X$containing$D_i$), cutting the surface into$R^2$again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into$R^2$. 3 added 133 characters in body Yes, this is true. Topologically, one may find a locally finite collection of properly embedded arcs in a connected surface whose complement is homeomorphic to$R^2$. Then make each of these arcs geodesic in the hyperbolic metric. The complement will be the fundamental domain of the type you want. Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a classification of connected surfaces. In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface$\Sigma\cong S^2-X$is obtained by removing a totally disconnected compact set$X\subset S^2$from$S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set$X\subset S^2$to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints)$X\subset I\subset S^2$. Then the properly embedded arcs$I\cap (S^2-X)$give a decomposition of$\Sigma$into$R^2$. If the surface is non-planar, then one removes from$S^2-X$a properly embedded countable collection of disks$D_1,D_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on$I$, and that the identifications either identifies antipodal points, or identifies two disks which are adjacent along a component of$I-X$. The complement$U=S^2-(I\cup_i D_i)$is again homemorphic to$R^2$. If we identify antipodal points of$D_j$, then this identifies two arcs in the boundary of$U$to obtain an open Mobius strip. We add two arcs connecting antipodal points of$D_j$to a point$x\in X$, at the end of the interval of$I-X$which intersects$D_j$, which forms a single arc after identification of antipodal points of$D_j$, and cuts the Mobius strip back up into$R^2$. If adjacent disks$D_i, D_{i+1}$are identified, then the complement of the arcs gives a thrice-punctured sphere. We add 4 arcs connecting these points to$x$, x$ (again, $x\in X$ is at the end of the interval of $I-X$ containing $D_i$), cutting the surface into $R^2$ again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into $R^2$.

2 added 1834 characters in body

Addendum: I'll add some comments on one way to obtain these properly embedded arcs in the infinite topology case. Ian Richards gave a classification of connected surfaces. In Theorem 3 of that paper, he explains how to construct all surfaces. A planar surface $\Sigma\cong S^2-X$ is obtained by removing a totally disconnected compact set $X\subset S^2$ from $S^2$. As explained in Prop. 5 of the paper, one may consider the totally disconnected set $X\subset S^2$ to be a subset of the Cantor set, and therefore a subset of the interval (including the endpoints) $X\subset I\subset S^2$. Then the properly embedded arcs $I\cap (S^2-X)$ give a decomposition of $\Sigma$ into $R^2$.

If the surface is non-planar, then one removes from $S^2-X$ a properly embedded countable collection of disks $D_1,D_2,\ldots$, and makes identifications of their boundaries. We may assume after an isotopy that these disks are all centered on $I$, and that the identifications either identifies antipodal points, or identifies two disks which are adjacent along a component of $I-X$. The complement $U=S^2-(I\cup_i D_i)$ is again homemorphic to $R^2$. If we identify antipodal points of $D_j$, then this identifies two arcs in the boundary of $U$ to obtain an open Mobius strip. We add two arcs connecting antipodal points of $D_j$ to a point $x\in X$, which forms a single arc after identification of antipodal points of $D_j$, and cuts the Mobius strip back up into $R^2$. If adjacent disks $D_i, D_{i+1}$ are identified, then the complement of the arcs gives a thrice-punctured sphere. We add 4 arcs connecting these points to $x$, cutting the surface into $R^2$ again. Continuing in this fashion inductively, we get a locally finite collection of arcs cutting the surface up into $R^2$.

1