If $X$ is a compact complex manifold, the exponential sequence gives an injective map $H^1(X,\mathbb{Z}) \to H^1(X,\mathcal{O}_X)$. I think that this shows that $H^1(X,\mathbb{Z})$ is torsion free.
Here is how a thought: Since $H^1(X,\mathcal{O}_X)$ is a vector space, if $n \alpha=0$ on $H^1(X,\mathbb{Z})$ then its image satisfies $n\alpha = 0$ on $H^1(X,\mathcal{O}_X)$ and hence $\alpha = 0$ on $H^1(X,\mathcal{O}_X)$ and consequently on $H^1(X,\mathbb{Z})$.
My question is wheter this argumet is right and if the result is true for every compact orientable manifold (ie, if every such manifold has torsionless $H^1$)