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).
