At least if $X$ is $\sigma$-compact and if compact sets have finite measure and if the local vector bundle trivializations are homeomorphisms, I would expect the representation of $(L^p(E))^*$ as $L^{p'}(E)$ to hold. The proof would proceed by first dividing $X$ into countably many disjoint relatively compact sets $B_i$ over which the bundle is trivial. Given $u$ in $(L^p(E))^*$, first consider the functionals $u_i:x\mapsto u(\bar x)$ on $L^p(E_{\,|B_i})$ where $\bar x$ denotes the zero extension. Then you get the representation of $u_i$ by some $y_i$ in $L^{p'}(E_{\,|B_i})$ . Patch these together to get a measurable section $y$ of $E$ such that $u(x)=\int_X\langle y(t),x(t)\rangle\,d\mu(t)$. Then you can show $\|y\|_{p'}\le\|u\|$ in the same way as this is done in the case of real or complex valued functions just by considering the inequality $|u(x)|\le\|u\|\,\|x\|_p$ for $x$ chosen so that it gets the form $\|y\|_{p'}^{\,p'}\le\|u\|\,\|y\|_{p'}^{\,\frac{p'}p}$ which directly gives the required inequality. Note here that you can get $\sum(y_ix_i)=(\sum y_i^2)^{\frac{p'}2}$ by choosing $x_i=\dfrac{y_i}{(\sum y_i^2)^{1-\frac{p'}2}}\ $.

At least if $X$ is $\sigma$-compact and if compact sets have finite measure and if the local vector bundle trivializations are homeomorphisms, I would expect the representation of $(L^p(E))^*$ as $L^{p'}(E)$ to hold. The proof would proceed by first dividing $X$ into countably many disjoint relatively compact sets $B_i$ over which the bundle is trivial. Given $u$ in $(L^p(E))^*$, first consider the functionals $u_i:z\mapsto u(\bar z_i)$ on $L^p(E_{\,|B_i})$ where $\bar z_i$ denotes the zero extension. Then you get the representation of $u_i$ by some $y_i$ in $L^{p'}(E_{\,|B_i})$ . Patch these together to get a measurable section $y$ of $E$ such that $u(x)=\int_X\langle y(t),x(t)\rangle_t\,d\mu(t)$ holds for $x$ in $L^p(E)$. Then you can show $\|y\|_{p'}\le\|u\|$ in the same way as this is done in the case of real or complex valued functions just by considering the inequality $|u(x)|\le\|u\|\,\|x\|_p$ for $x$ chosen so that it gets the form $\|y\|_{p'}^{\,p'}\le\|u\|\,\|y\|_{p'}^{\,\frac{p'}p}$ which directly gives the required inequality. Note here that you can get $\sum_{j=1}^n(\eta_j\xi_j)=(\sum_{j=1}^n\eta_j^2)^{\frac{p'}2}$ by choosing $\xi_j=\dfrac{\eta_j}{(\sum_{j=1}^n\eta_j^2)^{1-\frac{p'}2}}\ $.