If $S$ is any surface with $h^{1,0}(S) = 1$, then the Albanese morphism is a surjection $a \colon S \to E$ to an elliptic curve, and the pullback $H^0(E,\Omega_E^1) \to H^0(S,\Omega_S^1)$ is an isomorphism (see for instance [this post](https://mathoverflow.net/q/368825/82179)). So the only question is whether the pullback $a^* \omega$ of a nontrivial $1$-form $\omega$ on $E$ has a zero somewhere.

But as soon as $a$ has a nodal fibre (say), the pullback $a^* \omega$ vanishes at the node. For instance, you can argue that étale-locally, a family of curves with a nodal fibre looks like the map
$$V(xy-t) \subseteq \mathbf A^3_{(x,y,t)} \twoheadrightarrow \mathbf A^1_t.$$
Then the generator $\mathrm dt$ of $\Omega^1_{\mathbf A^1}$ pulls back to $\mathrm d(xy) = x\mathrm dy + y \mathrm dx$, which vanishes at $(0,0,0)$.

So any surface whose Albanese morphism is a surjection $a \colon S \to E$ with at least one nodal fibre gives a counterexample.

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P.S. If you want a more precise grip on numerical invariants of surfaces, there is a great tool by Pieter Belmans and Johan Commelin with loads and loads of data. See *Le superficie algebriche* at [https://superficie.info](https://superficie.info).