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Anton Petrunin
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Assume $M$ is complete. (It is easy to construct counterexamples in the noncomplete case.)

The inverse function theorem implies that the map $f$ is locally invertible; that is, if $p\in f(x)$, then there is a neighborhood $U\ni p$ and a right inverse $g\colon U\to M$ of $f$ such that $g(p)=x$. (Here we have to assume that $\varepsilon <1$.)

Now assume $\tilde \gamma=f\circ\gamma$ is a short loop based at $p$. Consider the geodesic homotopy $\tilde h_t$ from $\tilde \gamma$ to the constant map with image $p$. This homotopy admits a local lifting $h_t$ to $M$; moreover, the lengths of lifted curves $t\mapsto h_t(x)$ can be controlled. Therefore the maximal interval of definition of $h_t$ is a closed subinterval of $[0,1]$. On the other hand, this subinterval must be open. Hence $\tilde h_t$ admits a global lifting. In particular $\gamma$ is a loop --- a contradiction.

Anton Petrunin
  • 45k
  • 14
  • 135
  • 299