Maybe the following helps: Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces. There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.

• MR2343536; Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.

You can access the beginning of the book via scholar.google.com.

#Second Edit:

It seems to me now it is much simpler.

Proof: Measure theoretically, there are no nontrivial bundles. So you can find a global orthonormal frame by measurable sections $s_1,\dots,s_n$ of your bundle so that any section $f$ is of the form $f=\sum _i f^i.s_i$ where $(f^i)_{i=1}^n \in L^p(X,\mathbb R^n)$. This gives an isometry between $L^p$-sections of the bundle and a usual $\mathbb R^n$-valued $L^p$-space.

Maybe the following helps: Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces. There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.

• MR2343536; Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.

You can access the beginning of the book via scholar.google.com.

Second Edit:

It seems to me now it is much simpler.

Proof: Measure theoretically, there are no nontrivial bundles. So you can find a global orthonormal frame by measurable sections $s_1,\dots,s_n$ of your bundle so that any section $f$ is of the form $f=\sum _i f^i.s_i$ where $(f^i)_{i=1}^n \in L^p(X,\mathbb R^n)$. This gives an isometry between $L^p$-sections of the bundle and a usual $\mathbb R^n$-valued $L^p$-space.

Maybe the following helps: Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces. There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.

• MR2343536; Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.

You can access the beginning of the book via scholar.google.com.

#Edit:

You are right with your comment, it does not prove the converse. Let me try to sketch a proof (modelling on non-compact manifolds):
Let us assume that we have a countable trivializing cover $U_\alpha$ of $X$ and a subordinate "square partition of unity" $\phi_\alpha$ (so that $supp(\phi_\alpha)\subseteq U_\alpha$ for all $\alpha$, the family of supports is (locally finite does not make sense, what should I use?), $0\le \phi_\alpha(x)\le 1$, and $\sum_\alpha \phi_\alpha(x) =1$. Given $\lambda in L^p(X,E)^*$ and $f\in L^P(X,E)$, then $$\sum_{\alpha=1}^N \phi_\alpha . f \to f \text{ in } L^p \text{ for } N\to \infty,$$ (This is the key lemma and this needs to be true)
so that $\lambda(f) = \lim_N\sum^N \lambda(\phi_\alpha.f)$. Thus it suffices to show that $\lambda|_{L^p(U_\alpha,E)}$ is in $L^{p'}(U_\alpha, E)$ which is the trivialized version and thus true.

I hope you can fill the gaps (I know how to do them only on a smooth manifold).

Maybe the following helps: Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces. There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.

• MR2343536; Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.

You can access the beginning of the book via scholar.google.com.

#Second Edit:

It seems to me now it is much simpler.

Proof: Measure theoretically, there are no nontrivial bundles. So you can find a global orthonormal frame by measurable sections $s_1,\dots,s_n$ of your bundle so that any section $f$ is of the form $f=\sum _i f^i.s_i$ where $(f^i)_{i=1}^n \in L^p(X,\mathbb R^n)$. This gives an isometry between $L^p$-sections of the bundle and a usual $\mathbb R^n$-valued $L^p$-space.

Maybe the following helps: Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces. There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.

• MR2343536; Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.

You can access the beginning of the book via scholar.google.com.

Maybe the following helps: Theorem 3.12 (page 20) in the following source has such a related result, albeit for higher Sobolev spaces. There are quite subtle requirements for the trivialising atlas and the partition of unity which are used in the proof.

• MR2343536; Eichhorn, Jürgen Global analysis on open manifolds. Nova Science Publishers, Inc., New York, 2007. x+644 pp.

You can access the beginning of the book via scholar.google.com.

#Edit:

You are right with your comment, it does not prove the converse. Let me try to sketch a proof (modelling on non-compact manifolds):
Let us assume that we have a countable trivializing cover $U_\alpha$ of $X$ and a subordinate "square partition of unity" $\phi_\alpha$ (so that $supp(\phi_\alpha)\subseteq U_\alpha$ for all $\alpha$, the family of supports is (locally finite does not make sense, what should I use?), $0\le \phi_\alpha(x)\le 1$, and $\sum_\alpha \phi_\alpha(x) =1$. Given $\lambda in L^p(X,E)^*$ and $f\in L^P(X,E)$, then $$\sum_{\alpha=1}^N \phi_\alpha . f \to f \text{ in } L^p \text{ for } N\to \infty,$$ (This is the key lemma and this needs to be true)
so that $\lambda(f) = \lim_N\sum^N \lambda(\phi_\alpha.f)$. Thus it suffices to show that $\lambda|_{L^p(U_\alpha,E)}$ is in $L^{p'}(U_\alpha, E)$ which is the trivialized version and thus true.

I hope you can fill the gaps (I know how to do them only on a smooth manifold).