Since $\mathbb Q$ is not path-connected, I know that $H_0(\mathbb Q) \cong \bigoplus_{i=1}^\infty \mathbb Z$.
Also, $H_0(\mathbb Q) \cong \tilde H_0(\mathbb Q) \oplus \mathbb Z$.
So, is it true that $H_0(\mathbb Q) \cong \tilde H_0(\mathbb Q)$?
In other words, if $X \cong \bigoplus_{i=1}^\infty \mathbb Z$ and $X=Y \oplus \mathbb Z$, is $Y \cong \bigoplus_{i=1}^\infty \mathbb Z$?
The answer to your question about reduced homology is yes. The answer to your titular question is also yes: Since $Y$ is a direct summand of a free abelian group, it is a free abelian group. Additionally, we have that $\bigoplus\limits_{i \in \mathbb{N}} \mathbb{Z}$ is free on the set consisting of the basis of $Y$, lets call $B$, union a disjoint element. So the cardinality of $B \cup \{1\}$ is that of $\mathbb{N}$. Since the naturals are the lowest infinite cardinal, and clearly $B$ must be infinite, the cardinality of $B$ is that of the naturals which means $Y \cong \bigoplus\limits_{i \in \mathbb{N}} \mathbb{Z}$.
The argument can be easily adapted to show that that you can always solve for the isomorphism class of $Y$ from the isomorphism $F \cong Y \bigoplus \mathbb{Z}^n$, for finite $n$ if $F$ is a free abelian group where you know its rank.