"Why not simply use Bochner spaces like $C(X;L^2(\Omega))$?" --- do you mean that this would be the space of continuous sections of the bundle with fiber $L^2(\Omega)$? Yes, that is correct if the bundle is trivial, i.e., a bundle of the form $X \times H$ where $H$ is the fiber Hilbert space. But of course not all bundles have this form. A natural class of examples where the bundle is usually not trivial arises when $X$ is a Riemannian manifold and the fiber at a point is the tangent space there. Each tangent space carries an inner product, so this is a Hilbert bundle.
"Is the constant rank condition necessary in order to put a continuous structure on the bundle?" --- yes. It's hard to see how to reasonably topologize a bundle whose fibers could vary in dimension. One does sometimes consider "dimension drop" conditions where you only consider sections which, say, vanish at some point, or lie in a proper subspace of the fiber at some point.
"Is it true that the constant rank condition is not imposed on measurable Hilbert bundles because measurability is a weaker condition that allows for variation of fibers?" --- again, I basically agree. This setting is very different from the topological setting; measurable sections will be totally insensitive to any global topological features of a bundle. So one can just take a measurable Hilbert bundle over a measure space $X$ to be something of the form $\bigcup (X_n\times H_n)$, with $n$ ranging over $\mathbb{N} \cup \{\infty\}$, $(X_n)$ a measurable partition of $X$, and $H_n$ an $n$-dimensional Hilbert space. (The nonseparable setting introduces some bad pathology, so I prefer to stick to the separable case.)
While I'm on the subject, measurable Hilbert bundles provide a setting for spectral theory that nicely accomodates multiplicity:
$\bullet$ If $A$ is a (bounded or unbounded) self-adjoint operator on a Hilbert space $H$, then there is a measurable Hilbert bundle over the spectrum of $A$, and an isometric isomorphism between $H$ and the $L^2$ sections of this bundle which turns $A$ into multiplication by $x$.
$\bullet$ If $\mathcal{A} \subset B(H)$ is a separable abelian C*-algebra then there is a metrizable locally compact Hausdorff space $X$, a Borel measurable Hilbert bundle over $X$, and an isometric isomorphism between $H$ and the $L^2$ sections of the bundle which turns $\mathcal{A}$ into the set of multiplication operators by functions in $C_0(X)$.
$\bullet$ If $\mathcal{M} \subseteq B(H)$ is an abelian von Neumann algebra, then there is a metrizable locally compact Hausdorff space $X$, a Borel measurable Hilbert bundle over $X$, and an isometric isomorphism between $H$ and the $L^2$ sections of the bundle which turns $\mathcal{M}$ into the set of multiplication operators by functions in $L^\infty(X)$.
I would say that these statements cleanly exhibit the way the abstract C${}^*$- and von Neumann algebras are situated within $B(H)$.
(Details can be found in my book Measure Theory and Functional Analysis.)
