You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.
His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.
Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.
If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.
Added later: For an alternative proof, in a more analytical take, see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.