Yes, directly from the definition of fibration. And I see no advantage in assuming that the map is surjective. The fiber is empty if and if the homotopy fiber is empty.
I am guessing that your proof for $\pi_n$ with $n$ positive is a five lemma argument?
EDIT Now that I think about it, maybe my preferred proof goes by extending the 5 lemma argument a little. Like this: The homotopy fiber of $E\to B$ over $b\in B$ is the fiber of an associated fibration, call it $E'\to B$. There is a map $E\to E'$ over $B$ inducing a map of fibers $F\to F'$. You know that $E\to E'$ is a homotopy equivalence and so gives a bijection of $\pi_0$ and (for every basepoint in $E$) of $\pi_n$. To conclude that the associated map $F\to F'$ also induces such bijections, use a sort of modified 5 lemma argument. The key is that you have an action of $\pi_1(B,b)$ on $\pi_0(F)$ such that (a) for every point $f\in F$ the stabilizer of its class is the image of $\pi_1(E,f)$ and (b) two elements of $\pi_0(F)$ go to the same element of $\pi_0(E)$ if and only if they are in the same orbit.