Is it true that for every oriented vector bundle with odddimensional fibers, there is always a global section that vanishes nowhere?
No.
Let us consider some vector bundles over $S^4$. Since $S^4$ is simply connected all vector bundles are oriented and rank $k$ vector bundles over $S^4$ are classified by $\pi_{3}(SO(k))$ by the clutching construction (I am glossing over basepoint issues here, but I think it is correct in this setting). Now $\pi_3(SO(1))$ and $ \pi_3(SO(2)=\pi_3(S^1)$ are trivial groups, so there do not exist nontrivial rank $1$ and rank $2$ bundles over $S^4$. What about rank $3$? Well $SO(3)\cong \mathbb RP^3$ and the long exact sequence of homotopy groups of the fiber bundle $$ \mathbb{Z}_2\rightarrow S^3\rightarrow \mathbb{RP}^3 $$ shows that $\pi_3(SO(3))\cong\pi_3(\mathbb{RP}^3)\cong \mathbb{Z}$. Hence there are $\mathbb{Z}$ different rank $3$ vector bundles over $S^4$.
Only the trivial one admits a nonzero section. If another one admits a section, then the orthogonal complement (a rank $2$ bundle) must be nontrivial, otherwise the bundle is trivial. But our previous computation shows that there are nonontrivial rank $2$ bundles over $S^4$.

4$\begingroup$ Moreover an important example (often denoted $\Lambda^2_+$) of a rank $3$ oriented vector bundle with no nonvanishing section is given by the selfdual $2$forms on the round sphere $S^4$ (any other conformal structure and orientation would do). The reason is that otherwise $S^4$ would admit an (orthogonal) almost complex structure, hence two, one for each orientation ($S^4$ has an orentation reversing diffeo). Considering their only two (exercise) common tangent complex lines at each point, $TS^4$ would split in two rank $2$ subbundles, which is excluded as in this answer. $\endgroup$ – BS. Jul 22 at 10:28

$\begingroup$ Thanks a lot, it's wonderful! I conjectured it to be true in order to solve a exercise from Bott/Tu's GTM 82, "the Euler class of an oriented sphere bundle with evendimensional fibers is zero, when the sphere bundle comes from a vector bundle". I hope to know why this statement turn out to be true. $\endgroup$ – Fredy Jul 22 at 13:37

1$\begingroup$ @Fredy: Note that the euler class is also zero in this example. But this does not imply that there is a nonzero section. $\endgroup$ – Thomas Rot Jul 22 at 15:06

$\begingroup$ @ThomasRot: Thanks. Can you tell me more details about the proof of that exercise? I seem to lack other tools besides looking for a global section... $\endgroup$ – Fredy Jul 23 at 2:40

1$\begingroup$ @fredy: The map that sends each fiber $v$ to $v$ is orientation reversing if the vector space is odd dimensional. Hence $e(E)=e(E)$, so $2e(E)=0$, which in the de Rham theory implies that $e(E)=0$ (but not when the Euler class is defined with integral coefficients). $\endgroup$ – Thomas Rot Aug 6 at 9:42