Let $\pi: E \to B$ be a fibration of $E$ over $B$, let $F = \pi^{-1}(b)$ for some $b \in B$ be a representative fiber, and suppose that $B$ is contractible. Is it always the case (or are there some nice conditions guaranteeing) that $E$ is homotopic to $F$?
(For those curious, the context is John Milnor's work on the Milnor fibration: given an analytic function $f: \mathbf{C}^m \to \mathbf{C}$ with a singular point $f(0) = 0$ at the origin, the intersection of $f^{-1}(0)$ with a sufficiently small sphere $S_\epsilon^{2m-1}$ about the origin is transverse. Call this intersection $K$; the map $\pi(z) = f(z)/|f(z)|$ gives a locally trivial fibration of $S_\epsilon^{2m-1}-K$ over $S^1$. Milnor seems to rely implicitly on a result similar to the above.)
Yes, it is always the case. If you just need a weak homotopy equivalence, look at the long exact sequence in homotopy: $$\require{cancel} \dots \to \cancel{\pi_{n+1}(B)} \to \pi_n(F) \to \pi_n(E) \to \cancel{\pi_n(B)} \to \dots$$ to see that the inclusion $F \to E$ induces an isomorphism on all homotopy groups.
More generally, if you want a full-on homotopy equivalence, you can use the well-known fact that if $f_0, f_1 : B' \to B$ are homotopic and $E \to B$ is a fibration, then the two fibrations $f_0^* E \to B'$ and $f_1^* E \to B'$ are fibre homotopic. (The proof of this is a bit more technical.)
In the present case, let $f_0 = \operatorname{id}_B : B \to B$ and $f_1 : B \to B$ to be a constant map. Since $B$ is contractible, $f_0 \sim f_1$. Of course $f_0^*E = E$, while $f_1^*E \to B$ is the trivial fibration $B \times F \to B$. Since these two fibrations are fibre homotopy equivalent, their total spaces must be homotopy equivalent, i.e. $E \simeq F \times B$.
In fact we even see that there is a stronger result, that there is a commuting diagram where the vertical arrows are homotopy equivalences: $$\require{AMScd} \begin{CD} F @>>> E @>>> B \\ @V{\sim}VV @V{\sim}VV @V{=}VV \\ F @>>> B \times F @>>> B \end{CD}$$