Here is a counterexample for $G =\alpha_p$. Suppose that $k$ is a field of characteristic $p > 0$, and set $X = \mathbb A^2_k$, $Z= \{0\}$. Set $U := X \smallsetminus Z$. Then $\mathrm H^1(U, \mathbb{G}_{\rm a}) = \mathrm{H}^1(U, \mathcal{O})$ is an infinite dimensional vector space over $k$; hence it is $p$-torsion, and from the long exact sequence associated with the exact sequence
$$
0 \longrightarrow \alpha_p\longrightarrow \mathbb{G}_{\rm a}\longrightarrow \mathbb{G}_{\rm a}\longrightarrow 0
$$
we see that $\mathrm H^1(U, \alpha_p)$ surjects onto $\mathrm H^1(U, \mathbb{G}_{\rm a})$. If we take any class in $\mathrm H^1(U, \alpha_p)$ such that its image in $\mathrm H^1(U, \mathbb{G}_{\rm a})$ is $\neq 0$, this represents a torsor that does not extend to $X$, since $\mathrm H^1(X, \mathbb{G}_{\rm a}) = 0$.
[Edit:] Anon is right, my construction is nonsense. I apologize.