Yo! I was not able to find a counter-example to $$\tilde{H} (X/A) \cong H (X, A)$$.
It's a well known fact that for cofibrations $A \hookrightarrow X$ (or more generally whenever $A$ is a deformation retract of an open neighborhood (is this equivalent to being cofibration, by the way?)) the isomorphism holds.
The above isomorphism is equivalent to $$X \cup CA \cong X/A$$. So such space should be something so weird that the cone lying above $A$ does not retract to a point. I was thinking is something like a non-orientable surface embedded in a 3-dimensional space, like the Möbius band into $S^3$, however I was not able to make any considerable computation.
Another approach would be to see the complexes of abelian group directly. More precisely, let $f: C \rightarrow D$ be a map between two complexes. When $Cone (f)$ and $coker f$ are not quasi-isomorphic?
Thanks in advance
Let $X=[0,1]$ and $A=\{0,1,1/2,1/3,\cdots\}$. Then the quotient $X/A$ is homeomorphic to the Hawaiian earring which has uncountable $H_1$ (try to prove this). On the other hand, $H_1(X,A)$ is isomorphic to $\bigoplus_{i=1}^\infty \Bbb Z$, which is countable.