The idea you're looking for is called "fibrewise localization". It's defined in Dror Farjoun's book "Cellular Spaces, Null Spaces, and Homotopy Localization", and also in Hirschhorn's book (since you clearly have a copy of the latter, I'll use references from there).

Hirschhorn's Definition 6.1.1 defines the fiberwise localization of a map $p: E\to B$, of pointed spaces, as a factorization $p = q\circ i$ where $q$ is a fibration, and for every $b\in B$, the induced map on homotopy fibers $HFib_b(p) \to HFib_b(q)$ is a $\mathcal{C}$-local equivalence. In general, you construct such localizations by localizing the fibers. To be fibrant in the fiberwise localization model structure means that $p$ is a fibration and that all fibers are $\mathcal{C}$-local. When the fiberwise localization exists, it tells you how to replace your $F\to E\to B$ by the weakly equivalent $F\to E' \to B$.

Unfortunately, Hirschhorn's 6.1.4 proves that fiberwise localization DOES NOT EXIST in the category of pointed spaces. Further conditions are needed, as Kevin Carlson pointed out. So the general answer to your question has to be "no", or else fiberwise localization would exist for free.

One place that gives such conditions is Chataur and Scherer, "Fiberwise localization and the Cube Theorem", Theorem 4.3 (note that the "cube axiom" is required). In the situation of your question, $F$ is already local, so $\eta: F\to L_C(F)$ can be taken to be the identity. The proof of 4.3 shows that $E'$ (there denoted $\overline{E}$) is constructed as a localization of $E$, so is local, as you wanted.

everychoice of fiber. $\endgroup$ – Kevin Carlson Sep 14 at 21:03