You can look at Lieberman's paper [Holomorphic Vector Fields on Projective Manifolds][1].

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][2] 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.


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**Added later:** For an alternative proof see Theorem 6.4 of this [paper][3]. 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.


  [1]: http://books.google.com.br/books?id=pa1I1gG3eyUC&pg=PA273&lpg=PA273&dq=Holomorphic+vector+fields+on+projective+varieties.&source=bl&ots=QxbsauUatN&sig=8gFvKiAuur3UwDrP0eGPu94DLH4&hl=pt-BR&ei=KhIVTrmQKOPx0gGd-pRT&sa=X&oi=book_result&ct=result&resnum=4&ved=0CDgQ6AEwAw#v=onepage&q=Holomorphic%2520vector%2520fields%2520on%2520projective%2520varieties.&f=false
  [2]: http://www.jstor.org/pss/2372523
  [3]: http://arxiv.org/abs/1107.1538