I'm not aware of any generalization that is strong enough to compute the cohomology of such bundles completely, but you can at least use the Borel--Weil--Bott theorem to get some vanishing results. This was already done by Bott in his 1957 Annals paper to show that the cohomology of the tangent bundle $T$ of $G/P$ vanishes in degree > 0. The first step of this computation is to note that the action of $P$ on the tangent space at $P/P$ can be identified, as a $P$-module, with $\mathfrak g/ \mathfrak p$ (with $P$ acting via Ad). Consequently, the short exact sequence of $P$-modules $$ 0 \to \mathfrak p \to \mathfrak g \to \mathfrak g / \mathfrak p \to 0 $$ yields a short exact sequence of $G$-equivariant vector bundles $$ 0 \to G \times_P \ \mathfrak p \to G \times_P \ \mathfrak g \to T \to 0 $$ on $G/P$. The middle bundle is trivial, and so we get an isomorphism $H^q(G/P, T) \cong H^q(G/P, G \times_P \ \mathfrak p)$$H^q(G/P, T) \cong H^{q+1}(G/P, G \times_P \ \mathfrak p)$ for all $q>0$.
It remains to compute $H^q(G/P, G \times_P \ \mathfrak p)$$H^\ast(G/P, G \times_P \ \mathfrak p)$. For this, a vanishing result of the Borel--Weil--Bott type is useful. This result states that if you have a $G$-equivariant vector bundle $\mathcal V = G \times_P V$ on $G/P$ then $H^q(G/P, \mathcal V) = 0$ unless $q$ is equal to the index of some nonsingular $\mu + \rho$, where $\mu$ is one of the highest weights of $V$.
For the details, you can either try Bott's original paper, or checkout section 11 of this paper of Dennis Snow. Another paper of potential interest is
P.A. Griffiths, Some geometric and analytic properties of homogeneous complex manifolds. I. Sheaves and cohomology, Acta Math. 110 (1963), 115–155.