This is probably has an obvious proof or a straightforward counterexample, but I'm having trouble finding either.

Let $p:E \to B$ be a fibre bundle, with fibre $F$.  Assume that there is a spectrum $X$ and a homotopy equivalence of spectra

$$f: X \wedge \Sigma^{\infty} B_+ \to \Sigma^{\infty} E_+$$

(in particular, there is a stable section to $p$).  

Can we conclude that $X \simeq \Sigma^{\infty} F_+$?  If so, we may conclude that the bundle is stably trivial; that is, 

$$\Sigma^{\infty} E_+ \simeq \Sigma^{\infty} (B\times F)_+.$$

If this is not the case, what sort of conditions do we need to demand of the fibration to make it hold?  I'm happy to localize everything in site at your favorite prime, or for that matter, cohomology theory...

**Edit:**  There is an immediate type of counterexample, gotten by taking $E = B \times F$ to be trivial, but where $\Sigma^\infty F_+ \simeq X \vee Y$, where $Y$ is $\Sigma^\infty B_+$-acyclic; i.e., 

$$\pi_*(Y \wedge \Sigma^\infty B_+) = 0$$

One can define this problem away by assuming that $X$ and $\Sigma^\infty F_+$ are $\Sigma^\infty B_+$-local, which I'm happy to do for now.