EDIT. The answer is a resounding NO! ^^ there are obvious conter examples : take $A=B=\lbrace 0\rbrace$, and $X=\mathbb{R}\hookrightarrow\mathbb{R^2}=Y$, then $H_*(X,X-A)=H_*(\Bbb R,\Bbb R-0)=\Bbb Z[1]$ is $\Bbb Z$ in degree $1$, and $H_*(Y,Y-B)=H_*(\Bbb R^2,\Bbb R^2-0)=\Bbb Z[2]$ is $\Bbb Z$ in degree $2$. I feel silly !
Consider the following setup : I have a commutative diagram of spaces and maps $$ \begin{array}{ccc} B & \hookrightarrow & Y\\ \sim\uparrow\phantom{\sim} && \phantom{\sim}\uparrow\sim\\ A & \hookrightarrow & X\\ \end{array} $$ where $A\hookrightarrow X$ and $B\hookrightarrow Y$ are (Hurewicz) cofibrations, and the map of pairs $f:(X,A)\to(Y,B)$ induces a homotopy equivalence between $A$ and $B$, and between $X$ and $Y$.
Under these circumstances, it is for example known (prop 1.15 here) that the $f$ is a homotopy equivalence of pairs.
Is there necessarily an isomorphism of relative (co)homology $$ h(X,X-A)\simeq h(Y,Y-B) $$ ($h$ is singular (co)homology)
Of course there are immediate problems : while $f$ sends $A$ into $B$, it need not send $X-A$ into $Y-B$, although it could happen to do so in the case I'm considering, I haven't checked but it doesn't seem unreasonable.
If It can help, all spaces involved are Hausdorff and so $A$ and $B$ are closed subspaces of $X$ and $Y$ respectively, and most likely locally compact.