to compare cohomologies of fibers of two fiber bundles Consider the following commutative diagram of the fiber bundles $%
F\rightarrow E\rightarrow B$ and $F^{\prime }\rightarrow E^{\prime
}\rightarrow B^{\prime }$ where $B^{\prime }$ is simply connected space (but 
$B$ is not simply connected space) and all spaces are path-connected spaces.
$\require{AMScd}$
\begin{CD}
F @>{}>> E @>{}>> B \\ @VVV @VVV @VVV\\ F' @>{}>> 
E' @>{}>> B'
\end{CD}
Suppose that \begin{equation*}
H^{\ast }\left( B^{\prime };%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right) \rightarrow H^{\ast }\left( B;%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right) 
\end{equation*}
and 
\begin{equation*}
H^{\ast }\left( E^{\prime };%
%TCIMACRO{\U{211a} }
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right) \rightarrow H^{\ast }\left( E;%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right) 
\end{equation*}
are isomorphisms.
If 
\begin{equation*}
H^{i }\left( F;%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right)=0
\end{equation*} for all $i \geq n$ ($n$ fixed), then 
\begin{equation*}
H^{i }\left( F^{\prime };%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right)=0
\end{equation*}
for all $i \geq n$?
 A: No.  Let $B'$ be any space, and take $E'=PB'$ and $F'=\Omega B$.  The Kan-Thurston theorem gives a map $f\colon B\to B'$ such that $H^*(f;\mathbb{Q})$ is an isomorphism but $\Omega B$ is discrete, so $H^i(\Omega B;\mathbb{Q})=0$ for $i>0$.  The diagram
$\require{AMScd}$
\begin{CD}
\Omega B @>{}>> PB @>{}>> B \\
 @VVV @VVV @VVV\\ 
\Omega B' @>{}>> PB' @>{}>> B'
\end{CD}
satisfies most of your hypotheses, but $H^*(\Omega B')$ need not be bounded above.
One problem with the above example is that we have fibrations, but these need not be fibre bundles.  If necessary this can be fixed by a detour into simplicial sets and simplicial groups.
Another problem with the above example is that the space $F=\Omega B$ is disconnected.  If you want an example where absolutely everything is connected, we can proceed as follows.  We can assume that we have actual fibre bundles, and then let $\Sigma_BE$ denote the fibrewise unreduced suspension of $E$, which is a fibre bundle over $B$ with fibre $\Sigma F$, which is always connected.  We can describe $\Sigma_BE$ as the homotopy pushout of  $B\xleftarrow{}E\xrightarrow{}B$, and from this we see that the map $\Sigma_BE\to\Sigma_{B'}E'$ is a homology equivalence provided that $B\to B'$ and $E\to E'$ are homology equivalences.  Thus, we can apply this procedure to the previous counterexample to obtain a new counterexample in which everything is connected.
In general, if you know that a result is true for simply connected spaces, and you want to check whether that assumption can be relaxed, you should ask yourself whether the Kan-Thurston theorem gives counterexamples.
