It's easier to compute its $T$-equivariant $r$th Chern class $c_r(E) \in H^*_T(G/P)$, and use the maps $H^*_T(G/P) \twoheadrightarrow H^\ast(G/P)$ and $H_T^*(G/P) \hookrightarrow \oplus_{W/W_P} H_T^*(pt)$.
The latter comes from restricting the equivariant Chern class to the $T$-fixed points on $G/P$ (here $T$ is a maximal torus). The ring $H_T^*(pt)$ is the symmetric algebra on the weight lattice $T^\ast$, and $c_r(E|_{wP/P}) = w\cdot ($the $r$th elementary symmetric function of the $T$-weights on the vector bundle's base fiber, over $P/P \in G/P)$.
Now, how to show that's not zero, in ordinary cohomology... well, let's pull it back to $G/B$, first, and the cohomology injects so it's enough to work up there. On $\oplus_W H_T^\ast(pt) = [$functions $f : W \rightarrow H^\ast_T(pt)]$ there is a BGG/Demazure operator $\partial_\alpha$ for each simple root $\alpha$, taking $f$ to $w \mapsto (f(w) - f(w r_\alpha))/\alpha$. (It comes from pushing the class on $G/B$ to $G/P_\alpha$, then pulling it back.)
Your $c_r$ will be nonzero iff for some sequence of $r$ simple roots $\alpha_1,\ldots,\alpha_r$, this $\partial_{\alpha_1}\cdots \partial_{\alpha_r} c_r$ (now necessarily a constant function on $W$) is not zero.
Not knowing your vector bundle, I can't advise you what sequence of simple roots to use, other than that it should give a reduced word in $W$. EDIT: you can choose them greedily, applying $\partial_\alpha$ and trying to stay nonzero. If you get to a point (before getting to a constant) where all $\partial_\alpha$ give zero, then your Euler class is zero.
One place this combinatorial approach to equivariant cohomology of $G/P$ is laid out is my paper with Terry Tao on the equivariant cohomology of Grassmannians.
