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](http://www.ams.org/mathscinet-getitem?mr=2343536); 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.