Yes, and we do not need to assume that $\mathcal{G}$ is locally free (this happens automatically). The question is local, so we reduce immediately to the case that $X = \text{Spec } A$ is affine and $\mathcal{F}$ corresponds to a projective $A$-module. A direct summand of a projective module is certainly projective, since a module is projective if and only if it is a direct summand of a free module.
Edit: As Zhen Lin points out, there is something slightly nontrivial happening here. Namely, in order for this argument to work, we must assume that $\mathcal{F}$ is coherent. The fact that I am using is that a finitely presented module is projective if and only if it is locally free.