An example is the Möbius bundle $E \to S^1$. The Thom space is $\mathbb RP^2$ which does not have the same $\mathbb Z$ cohomology (modulo a shift by 1) as $S^1$ since $H^*(\mathbb R P^2)$ has torsion. To see that $\mathbb R P^2$ is the Thom space, note that the unit disk bundle is a rectangle with one pair of opposite ends identified with a twist (the closed Möbius strip). Then identifying all of the edges to a point amounts to gluing the 2-ball to itself via the antipodal map on the boundary, which gives $\mathbb R P^2$.
As for your real question, orientability is equivalent to the existence of an element $u \in H^n(E, E_0)$ that restricts to a generator on each fiber. In the case of the trivial vector bundle you are using a choice of orientation. In general this won't be possible. If you think in terms of finding a non-zero section of $\Lambda^n E$, $n = rk~E$, you can always do this locally as $e_1 \wedge \cdots \wedge e_n$, where $e_1,\ldots,e_n$ is an orthonormal basis. If you chose a different orthonormal basis (e.g. a different trivialization) then this can change by $\pm 1$, and there is a topological restriction to making this choice consistently. It would be a good exercise to work this out on the Möbius band (where you can take a cover where the transition function is just multiplication by -1) to see why there doesn't exist a global top form (here just a single non-vanishing section and for real line bundles, orientability is equivalent to being trivial).