Let S be a hyperbolic surface. Let $\phi : B_{S}(x,R) \rightarrow B_{H}(\phi(x),R)$ is an isometry from a ball in S to a ball in hyperbolic plane. I claim that there is no locally minimizing closed geodesic in $B_{S}(x,R)$.
This is my idea: Suppose that there is a locally minimizing closed geodesic in $B_{S}(x,R)$, namely $\gamma$. Then $\phi\circ\gamma$ is a locally minimizing closed geodesic in $B_{H}(\phi(x),R)$. I know that is a contradiction because there is no closed geodesic in $H$ but I can not write down a detail proof for it. Hoping someone can help me! Thank in advance !
You need to assume that $R$ is less than the injectivity radius of $S$. If $R$ is greater than the injectivity radius, then $B_S(x,R)$ is not simply connected and there is no isometry from $B_S(x,R)$ to a ball in the hyperbolic plane.
If $R$ is less than the injectivity radius of $S$, then your argument holds. It would work to appeal to the classification of geodesics in one of the common models of the hyperbolic plane, e.g., geodesics of the Poincare disk model of $H$ are exactly the Euclidean circles which are perpendicular to the unit circle.