$\newcommand\th\theta\newcommand\la\lambda\newcommand\R{\mathbb R}$Note that for $t\in\R^d$ we have
$$M_\th(t):=E_\th e^{t\cdot T(X)}
=\int_{\R^d}dx\,h(x)e^{(t+\th)\cdot T(x)-\phi(\th)}
=e^{\phi(t+\th)-\phi(\th)},$$
where $\cdot$ denotes the dot product. So,
$$E_\th T(X)=\nabla M_\th(0)=\nabla\phi(\th)$$
and, for any real $\la$,
$$E_\th e^{\la(Z-E_\th Z)}
=E_\th\exp\{\la v\cdot T(X)-\la v\cdot E_\th T(X))\} \\
=\exp\{\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)\}.$$$$E_\th e^{\la(Z-E_\th Z)}
=E_\th\exp\{\la v\cdot T(X)-\la v\cdot E_\th T(X))\} \\
=\exp\{\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)\}.
\tag{1}\label{1}$$
By the mean-value theorem, for some $a\in(0,1)$ depending on $\la,v,\th$, we have
$$\phi(\la v+\th)-\phi(\th)=\la v\cdot\nabla\phi(a\la v+\th).$$
Also, by the Lipschitz condition on $\nabla\phi$,
$$|\nabla\phi(a\la v+\th)-\nabla\phi(\th)|\le La\la|v|\le L\la.$$
Thus,
$$\phi(\la v+\th)-\phi(\th)-\la v\cdot \nabla\phi(\th)
=\la v\cdot(\nabla\phi(a\la v+\th)-\nabla\phi(\th))
\le L\la^2$$
and hence, by \eqref{1},
$$E_\th e^{\la(Z-E_\th Z)}
\le e^{L\la^2},$$
so that $Z$ is (uniformly) sub-Gaussian for all $\th$.
