$\DeclareMathOperator{\Ext}{Ext}$The exact sequence mentioned in Wikipedia is only a corollary of a more general theorem; the corollary is only true for PIDs. The spectral sequence has the form:
$$E_2^{p,q} = \Ext{}^q_S(H_p(X), R) \Rightarrow H^*(X; R)$$
I won't delve into the details if you don't know anything about spectral sequences, but when $S$ is a PID (so for example $S = \mathbb{Z}$) the higher $\Ext$ functors disappear (because every module has a projective resolution of length at most one). So the spectral sequence degenerates, and you're left with the exact sequence that you mention. But in full generality, you cannot get rid of the higher $\Ext$ terms, the differentials that appear in the SS, and the extension problems at the $E_\infty$ page.
Here is the situation. $\require{AMScd}$ The statement that the universal coefficient theorem is natural for maps $f: C \to D$ of chain complexes of free abelian groups means that there is a map of short exact sequences
$\begin{CD}0 @>>> \text{Ext}^1_{\Bbb Z}(H_{i-1}(C), G) @>>> H^i(C;G) @>>> \text{Hom}(H_i(C), G) @>>> 0\\
@. @AAA @AAA @AAA \\
0 @>>> \text{Ext}^1_{\Bbb Z}(H_{i-1}(D), G) @>>> H^i(D;G) @>>> \text{Hom}(H_i(D), G) @>>> 0
\end{CD}$
Note that the arrows go the other way from $f: C \to D$; all three of $\text{Ext}(-,G)$, $\text{Hom}(-, G)$, and cohomology with $G$ coefficients are contravariant.
Because $\text{Ext}$ and $\text{Hom}$ are functorial in their inputs, they take isomorphisms to isomorphisms; in particular, because $f_*: H_k(C) \to H_k(D)$ is an isomorphism for any $k$, we see that the outer two vertical maps are isomorphisms.
Whenever you have a big diagram like this and some of those vertical arrows start being isomorphisms, you should think about the 5 lemma. In this case, though it looks like we only have 3 vertical arrows, don't forget about the implicit maps $0 \to 0$ on the outside (the trivial map!). This means that we have a map of 5-term exact sequences, which are isomorphisms on the "outer 4" vertical maps. Then the 5 lemma gets us that the middle vertical map is an isomorphism; this is what you wanted.
Best Answer
Here is an answer assuming that $\alpha$ is a chain map; without this assumption I do not see how you can talk about induced maps of integer homology groups. Then the answer to your question is positive and follows from the Universal coefficient Theorem (UCT). Indeed, since $\alpha$ is a chain map, it induces maps $\alpha_{i*}: H_i(C; G)\to H_i(C'; G)$ for arbitrary abelian groups $G$. Since the statement of the UCT provides a natural short exact sequence, we get isomorphisms $$ \beta_i: H_i(C)\otimes G \to H_i(C')\otimes G $$ $$ \gamma_i: Tor(H_{i-1}(C), G)\to Tor(H_{i-1}(C'), G) $$ which (together with $\alpha_{i*}$) make a commutative diagram with UCT sequences for $C, C'$ on the top and on the bottom. (Sorry, I am not very good at drawing diagrams in the version of TeX used at MSE.) Now, the fact that $\alpha_{i*}$ is an isomorphism follows from the diagram chasing in this commutative diagram. For instance, to see injectivity, note that an element $x$ of the kernel has to project to zero in $Tor(H_{i-1}(C), G)$. Hence, it belongs to $H_i(C; {\mathbb Z})\otimes G$, but then injectivity of $\beta_i$ implies that $x=0$.