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). 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. For instance, if $S$ is a surface of general type with $h^{1,0}(S) = 1$ (in this case, the vanishing of its global $1$-form also follows from [PS14]).

References.

[PS14] M. Popa, C. Schnell, Kodaira dimension and zeros of holomorphic one-forms. Ann. Math. (2) 179.3 p. 1109-1120 (2014). ZBL1297.14011.

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.

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). So the only question is whether the pullback $a^* \omega$ of a nontrivial $1$-form $\omega$ on $E$ has a zero somewhere.

If $a$ is not smooth, then the pullback $a^* \omega$ vanishes at any singular point $s \in S$ for $a$. Indeed, note that $a$ is singular at $s$ if and only if $\dim_{\kappa(s)}\bigl(\Omega^1_{S/E} \otimes \kappa(s)\bigr) > 1$, which means that the pullback $a^* \colon \Omega^1_E\otimes \kappa(a(s)) \to \Omega^1_S \otimes \kappa(s)$ is the zero map (since $\dim_{\kappa(s)} \Omega_S^1 \otimes \kappa(s) = 2$).

So any surface whose Albanese morphism is a surjection $a \colon S \to E$ with at least one singular fibre gives a counterexample. For instance, if $S$ is a surface of general type with $h^{1,0}(S) = 1$ (in this case, the vanishing of its global $1$-form also follows from [PS14]).

References.

[PS14] M. Popa, C. Schnell, Kodaira dimension and zeros of holomorphic one-forms. Ann. Math. (2) 179.3 p. 1109-1120 (2014). ZBL1297.14011.

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.

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). 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.

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.

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). 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. For instance, if $S$ is a surface of general type with $h^{1,0}(S) = 1$ (in this case, the vanishing of its global $1$-form also follows from [PS14]).

References.

[PS14] M. Popa, C. Schnell, Kodaira dimension and zeros of holomorphic one-forms. Ann. Math. (2) 179.3 p. 1109-1120 (2014). ZBL1297.14011.

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.