This is only a partial answer, but it is not true in general even for $K3$ surfaces. Let $S$ be a smooth algebraic $K3$, containing a smooth rational curve $C \stackrel{\gamma}{\hookrightarrow} S$. Then by adjunction, the normal bundle is $\mathcal{N}_{C|S} \cong \mathcal{O}_C(-2)$. There is a short exact sequence
$$
0 \to \mathcal{N}_{C|S}^{\check{}} \to \gamma^*\Omega^1_S \to \Omega^1_C \to 0 ~,
$$
which becomes
$$
0 \to \mathcal{O}_C(2) \to \gamma^*\Omega^1_S \to \mathcal{O}_C(-2) \to 0 ~.
$$
Since $\mathcal{O}_C(2)$ doesn't inject into $\mathcal{O}_C^{\oplus 2}$, $\gamma^*\Omega^1_S$ must be non-trivial.
As suggested by Tom, the above can be generalised (and it can probably be generalised further by somebody who knows their stuff):
If $X$ is a smooth variety containing a smooth rational curve $\gamma : C \hookrightarrow X$, then $X$ does not have the property given in the OP. To see this, start with the following exact sequence:
$$
0 \to {N}_{C|X}^{\check{}} \to \gamma^*\Omega^1_X \to \Omega^1_C \to 0
$$
where $N_{C|X}^{\check{}}$ is the conormal bundle. Taking determinants, we find
$$
\gamma^* \omega_X \cong \omega_C\otimes\det (N_{C|X}^{\check{}}) \cong \mathcal{O}_C(-2)\otimes\det (N_{C|X}^{\check{}})
$$
So if $\gamma^*\Omega^1_X$ is trivial, we must have $\det (N_{C|X}^{\check{}}) \cong \mathcal{O}_C(2)$. As $N_{C|X}^{\check{}}$ is isomorphic to a sum of line bundles, at least one of them must have positive degree. But then it cannot inject into a trivial bundle, so we have a contradiction.