Here are three special cases where it is always true:
Sphere bundles of vector bundles. If $E \to B$ is a vector bundle and $S(E) \to B$ is the associated sphere bundle, then $S(E)$ bounds the disc bundle $D(E)$, so it is nullcobordant. On the other hand, $S^{n-1}\times B$ is nullcobordant as $S^{n-1}$ is, so $S(E)$ and $S^{n-1}\times B$ are cobordant. Provided $E$ is an orientable bundle, the same is true in oriented cobordism.
Finite Covering spaces. If $F \to E \xrightarrow{\pi} B$ is a finite covering space (i.e. $F$ is finite), then $TE \cong \pi^*TB$. It follows that the Stiefel-Whitney numbers of $E$ are just the Stiefel-Whitney numbers of $B$ multiplied by $\deg \pi = |F|$. So $E$ has the same Stiefel-Whitney numbers as the disjoint union of $|F|$ copies of $B$ (Stiefel-Whitney numbers add under disjoint sum), which is nothing but $F\times B$.
A similar consideration with Pontryagin numbers shows that if $E$ and $B$ are oriented, then the same conclusion holds in oriented cobordism.
- Principal bundles. If $G \to E \xrightarrow{\pi} B$ is a principal $G$-bundle, and $\dim G \geq 1$, then $E$ is nullcobordant. To see this, note that $TE \cong V\oplus H$ where $V$ and $H$ are the vertical and horizontal spaces respectively. Now $V\cong E\times\mathfrak{g}$ while $H \cong \pi^*TB$, so $TE \cong (E\times\mathfrak{g})\oplus\pi^*TB$. In particular, $$w(TE) = w((E\times\mathfrak{g})\oplus\pi^*TB) = w(\pi^*TB) = \pi^*w(TB).$$ As $\dim B < \dim G$, all the Stiefel-Whitney numbers of $E$ vanish, so it is nullcobordant. On the other hand, $G\times B$ is nullcobordant because $G$ is (Lie groups are parallelisable, so all their Stiefel-Whitney numbers are zero). Again, provided that $E$ and $B$ are orientable, the same is true in oriented cobordism (the Pontryagin numbers of $E$ are also zero).