I believe what you are asking is true whenever $X\to B$, considered as an unequivariant map,
is a Serre fibration.

** First some definitions: **

Call a map of $G$-spaces $E \to B$ a $G$-Serre fibration if and only if for all subgroups $H\subset G$, the map of fixed points $E^H \to B^H$ is a Serre fibration. In particular,
$EG \times E \to EG \times B$ is a $G$-Serre fibration if and only if it is a Serre fibration of unequivariant spaces. It is known this notion of fibration arises from a model structure on $G$-spaces, in which a map $X\to Y$ is a $G$-weak equivalence iff each map of fixed point sets $X^H \to Y^H$ is a weak homotopy equivalence. A map $X\to Y$ is a $G$-cofibration iff $Y$ is obtained from $X$ by attaching cells of the form $D^n \times (G/H)$ where $H$-varies through subgroups and the attaching maps are $G$-maps, or more generally if the pair $(Y,X)$ is a retract of a relative $G$-cell complex.

I do not know a reference for the above, but I am confident it's in the literature.
(Added Later: see the comment below for two references.)

** Now the argument: **

Suppose that $A\to U$ is an acyclic cofibration in the Serre model structure on spaces. Without loss in generality, we can assume that $U$ is obtained from $A$ by cell attachments. Suppose we are given a lifting problem:

$A \to X \times_G EG $

$\downarrow\qquad \qquad \downarrow$

$U \to B\times_G EG$

* We need to find a map $U \to X\times_G EG$ making the diagram commute. *

Here's how: pull back the above to an equivariant lifting problem

$\tilde A \to X \times EG $

$\downarrow\qquad \qquad \downarrow$

$\tilde U \to B\times EG$

where $\tilde A$ for example is given by the pullback of $A \to BG \leftarrow EG$
(the map $A\to BG$ is the composite $A\to X\times_G EG \to BG$). It is relatively straightforward to check that the inclusion $\tilde A\to \tilde U$ is an acyclic $G$-cofibration,
where the cells that are being attached are of the form
$D^n \times G$, i.e., they're free.

It follows from the model category structure on $G$-spaces that there's an equivariant lift $\tilde U \to X\times EG$ making the diagram commute. Now take orbits to get the a lift
$$
U \to X\times_G EG
$$
solving the original lifting problem.