No, there need not be such a geometric triangulation.
Construction: Consider $A$, a flat annulus, of width $W$ and length $L$. Here we assume that $W$ is very large and $L$ is very small. (That is, take a $W$ by $L$ rectangle and glue the long sides.)
Let $\alpha$ and $\beta$ be the components of $\partial A$. We glue many sub-intervals in $\alpha$ to isometric sub-intervals in $\beta$, via some complicated permutation. This gives a high genus surface $S$. The singular points all live in a graph -- namely the image of $\partial A$ after taking the quotient. (In fancier language: $S$ is the suspension of an interval exchange transformation.)
Proof: Note that the diameter of $S$ is at most $(W+L)/2$. However, any geometric triangulation must have at least two edges crossing $A$ and these edges have length at least $W$. So we are done.
$\newcommand{\diam}{\mbox{diam}}$I actually I tried to prove that the answer was "yes" before realizing that there are counterexamples. One natural way provide a "yes" answer is via the Delaunay triangulation. This lead me to Theorem 4.4 of the paper "Hausdorff dimension of sets of nonergodic measured foliations" by Masur and Smillie. They define $d(S)$ to be the maximal distance of a non-singular point $x \in S$ to the set of singularities $\Sigma \subset S$. Note that $d(S) \leq \diam(S)$. Their Theorem 4.4 says that all of the Delaunay polygons in $S$ can be inscribed in a disk of radius at most $d(S)$. Thus the length of any Delaunay edge is at most $2 \cdot d(S)$.