The answer is yes:

Let me sketch the proof. So $p:E\to M$ is the fiber bundle with typical fiber $F$ which is compact, connected (and oriented, for simplicity's sake), and you are given a vertical volume form $\mu$; so $\mu_x$ is a volume form on each fiber $E_x$ which depends smoothly on $x\in M$. First I choose another vertical volume form $\nu$ such that the volume of each fiber is 1, $\int_{E_x} \nu_x=1$. Take $\nu_x = \frac{\mu_x}{\int_{E_x}\mu_x}$, for example.

Now I construct the Hilbert bundle with fibers $L^2(E_{x},\nu_{x})$:
Fix a Riemannian metric $g$ on $F$ with $\int_F vol(g)=1$.
Let $U\subset M$ be open so that $\phi:U\times F \to E|U$ is a fiber respecting diffeomorphism.
For each $x\in M$ the Moser trick gives us a diffeomorphism $\psi_x:F\to F$ depending smoothly on $x\in U$ with $(\psi_x)^*(\phi_x)^*\nu_x = vol(g)$. This uses the Green function of the Hodge decomposition with respect to $g$ to choose a $(\dim(F)-1)$-form $\alpha_x$ with $d\alpha_x = \phi_x^*\nu_x-vol(g)$ which depends still smoothly on $x\in U$.

Edit: 43.7 in the book cited below contains Moser's trick in the form I just described.

Then the mapping $\bigsqcup_{x\in U}(x, L^2(E_{x},\nu_{x}))\ni (x,f) \mapsto (x,f\circ \phi_x \circ \psi_x^{-1})\in U\times L^2(F,vol(g))$
is an isometric trivialisation of the bundle
$\bigsqcup_{x\in M}(x, L^2(E_{x},\nu_{x}))$ over $U$.

Edit (more details):
The change of trivialisation is then of a similar form, $(x,f)\mapsto (x,f\circ \rho_x)$
for smooth $\rho:U\times F\to F$ such that $\rho_x$ is a $vol(g)$-preserving diffeomorphism for each $x\in U$.
That it is smooth $U\times L^2(F, vol(g)) \to U\times L^2(F,vol(g))$ is seen as follows:
It suffices to show that $(x,f)\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2}$ is smooth for all $\lambda$ in a subset $\subset L^2$ of linear functionals which together recognize bounded sets.
We may take $C^\infty(F)\subset L^2(F,vol(g))$ as this set. By one of the two smooth uniform boundedness theorems from the book below it suffices to show that for each fixed $f\in L^2$ the function $F\to \mathbb R$ given by
$$x\mapsto \langle f\circ \rho_x, \lambda\rangle_{L^2} = \int_F f(\rho_x(u))\lambda(u)\,vol(g)(u)= \int f(v) \lambda(\rho_x^{-1}(v) ((\rho_x^{-1})^*vol(g))(v)$$
is smooth.
But this now obvious since $\lambda$ and $vol(g)$ are smooth.

The original inner product $\int_{E_x} f \mu_x$ is now a fiberwise Riemann metric on this Hilbert bundle.

I use calculus in infinite dimensions from:
Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997,
(pdf).

# Edit:

As TaQ noted in his answer, my proof above is wrong. In fact, the answer is no, if you accept that the construction which I tried is the natural one. Namely, in the realm of Sobolev spaces, if $k>\frac{\dim(F)}2$, for the composition mapping $H^{k+l}(F,\mathbb R) \times H^k(F,F) \to H^k(F,\mathbb R)$, left translations are $C^l$ and right translations are smooth; i.e., composition is $C^l$ in the right hand side variable, and is smooth in the left hand side variable. This is folklore; for a detailed proof see

- H. Inci,T. Kappeler and P. Topalov, On the Regularity of the Composition of Diffeomorphisms, Memoirs of the American Mathematical Society, vol. 226 (American Mathematical Society, 2013).

In the case above we have left translations, and no assumption for to be above the Sobolev threshold.

But if one asks for Sobolev spaces instead of $L^2$, one gets a $C^{k}$ vector bundle for $H^{m}$ with $m> k + \frac{\dim(F)}2$.

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