To define an ideal triangulation one typically starts with a hyperbolic metric. Suppose that $T^2 = S^1 \times S^1$ is the two-torus and $X = T - D$ is the torus minus an closed embedded disk. There are essentially two ways to give $X$ a complete hyperbolic metric. As everyone has remarked, and you have remarked as well, if the metric has finite area then you can triangulate $X$ with exactly two ideal triangles with all vertices at the "ideal point at infinity."

On the other hand, the other "kind" of metric on $X$ is a metric of infinite area. Since an ideal triangle has area $\pi$ it will take infinitely many ideal triangles to triangulate $X$. These triangulations can be very wild. In particular they are not typically determined by a finite amount of data. Thus the phrase "ideal triangulation" virtually always refers to the case when $X$ has finite area.

EDIT: Of course I misread the question! The question asks about ideal triangulations of a hyperbolic surface with geodesic boundary. Here the answer pretty much has to be "spun triangulations". That is: fix such a metric on $X = T - D$ when $D$ is an open disk. Fix a collection of three proper embedded geodesic arcs in $X$, perpendicular to the geodesic boundary. These cut $X$ into a pair of right-angled hexagons. Now we spin. Choose an orientation for $\partial X$. Drag all of the endpoints of all the arcs in the direction of the orientation. At each point of time during this process we have three proper geodesic arcs, but the angles they make with the boundary go to zero. Take the Hausdorff limit and you will get a spun ideal triangulation of $X$. I'll end by remarking that you can build a spun triangulation of a pair of pants ($S^2$ minus three open disks with disjoint closures).