Let us consider a homotopy fibre sequence of connected spaces $A\rightarrow B\rightarrow C$ and let $K$ be a fixed field. Assume that the homology $H_{\ast}(A, K)$ is trivial and that $C$ is a nilpotent space (but not simply connected).
Does $H_{\ast}(B, K)\rightarrow H_{\ast}(C, K)$ have to be an isomorphism ?
In your situation, the Serre spectral sequence looks as follows
\[H_p(C,H_q(A,K))\Rightarrow H_{p+q}(B,K).\]
The left hand side is the $E^2$ term. There, $H_q(A,K)$ carries an action of $\pi_1(C)$ induced by the fibration, and $H_p(C,H_q(A,K))$ is the homology with local coefficients. Now, if $A$ has the homology of a point then then $H_q(A,K)=0$ for $q>0$ and $H_0(A,K)=K$ with the trivial action of $\pi_1(C)$ (any self-map of a connected space induces the identity on $H_0$). Therefore $E^2_{p,q}=0$ for $q>0$ and $E^2_{n,0}=H_n(C,K)$ is the ordinary homology. Moreover, the spectral sequence ends in the second step and the edge morphism $H_n(B,K)\twoheadrightarrow E^\infty_{n,0}\subset E^2_{n,0}=H_n(C,K)$, which is induced by the map $B\rightarrow C$, is an isomorphism.
Here is a compromise between Fernando's answer and S. Carmeli's comment. Also, we may as well use $E_*$ any homology theory and there's no need to assume $C$ is nilpotent, just that every fiber is $E_*$-acyclic (so, e.g., if $C$ is connected and you have your assumption then we're okay.)
The statement is okay with filtered colimits and equivalences, so we may as well assume $C$ is a finite dimensional CW-complex and induct on the dimension. For the inductive step, observe that $\mathrm{sk}_{n-1}C \to \mathrm{sk}_nC$ is obtained by cobase change from $\coprod \partial D^n \to \coprod D^n$, and it follows that $B\vert_{\mathrm{sk}_{n-1}C} \to B\vert_{\mathrm{sk}_{n}C}$ is obtained by cobase change along $\coprod(B\vert_{\partial D^n} \to B\vert_{D^n})$. These two pushout diagrams are homotopy pushouts, and the projection map from one to the other is an $E_*$-equivalence on the cospans by induction and the assumption on the fiber since $B\vert_{D^n} \to D^n$ is equivalent to $A \to \bullet$ after trivializing the fibration.
(This is a compromise of S. Carmeli's answer because you can compute the `homotopy colimit over $C$' by decomposing $C$ along a cell diagram and computing the colimit one piece at a time; this is a compromise of Fernando's answer since one often constructs the Serre spectral sequence using the above pushout diagrams anyway.)
Yet another rephrasing. Homology/chains can be computed in terms of the functoriality of local systems of complexes. In this language, any homotopy type $C$ has a constant local system $k_C$, and the chains $C_{*}(C,k)$ are computed as $p_!(k_C)$, where $p_!$ is the left adjoint to the pullback functor of local systems for $p: C \rightarrow *$. The important, relevant fact is that local systems satisfy base-change for homotopy fibre products, such as your fibre sequence. For connected base $C$, your hypothesis that the homology of the fibre is trivial is equivalent to the hypothesis that the canonical map $f_{!}(k_B) \rightarrow k_C$ of local systems is an isomorphism, since the fiber of $f_{!}(k_B)$ is just $k$, by base-change and the hypothesis on the homology of the fibre $A$. Now apply $p_{!}$ to $f_{!}(k_B) \rightarrow k_C$ to obtain the isomorphism $C_{*}(B,k) \rightarrow C_{*}(C,k)$.
(Remarks. Really, such an argument has to take place $\infty$-category land, or at least in derived categories. Also, the same argument works for any other homology theory, by changing the system of coefficients.)
