Here are some comments about the case where $X$ is not assumed to have finite type. Put $Y=\Sigma X$. For any field $K$, the groups $H_*(Y;K)$ form a Hopf algebra in which all elements of the augmentation ideal are primitive. If $u$ and $v$ lie in the augmentation ideal, then $u$, $v$ and $uv$ are all primitive, which gives $u\otimes v+(-1)^{|u||v|}v\otimes u=0$. If $u$ and $v$ are nonzero, it follows that $|u|=|v|=1\pmod{2}$ and $Ku=Kv$. Thus, $H_*(Y;K)$ is either $K$ or $K\oplus Ku$ for some element $u$ of odd degree. In particular, we see that $Y$ is rationally trivial or an odd-dimensional sphere.

Now consider the groups \begin{align*} A(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}_{(p)}) \\ B(p)_k &= A(p)_k/p \\ C(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}/p) \\ D(p)_k &= \text{ann}(p,A(p)_{k-1}) \end{align*} so that

• $A(p)\otimes\mathbb{Q}$ has dimension $0$ or $1$ over $\mathbb{Q}$
• $C(p)$ has dimension $0$ or $1$ over $\mathbb{Z}/p$
• There is a short exact sequence $B(p)\to C(p)\to D(p)$, so $B(p)$ and $D(p)$ have dimension $0$ or $1$, and at least one of them is zero.

There are a number of different possibilities here.

• If $D(p)=0$ then $A(p)$ is torsion-free and so injects in $A(p)\otimes\mathbb{Q}$. This means that $A(p)=0$ or $A(p)\simeq\mathbb{Z}_{(p)}$ or $A(p)\simeq\mathbb{Q}$.
• If $B(p)=0$ then $A(p)$ is divisible, and so is injective as a $\mathbb{Z}_{(p)}$-module. If we let $T(p)$ denote the torsion part of $A(p)$ then we find that $T(p)$ is also divisible and therefore injective and therefore a summand in $A(p)$. We therefore have $A(p)=T(p)\oplus Q(p)$, where $Q(p)$ is a $\mathbb{Q}$-module. It follows that $A(p)\otimes\mathbb{Q}\simeq Q(p)$, so $Q(p)$ is $0$ or $\mathbb{Q}$. If $D(p)=0$ then $T(p)=0$. If $D(p)=\mathbb{Z}/p$ then I think it follows that $T(p)=\mathbb{Z}/p^\infty$.

The most interesting question arising from this analysis is as follows. Let $Y$ be the Moore space with $H_2(Y)=\mathbb{Z}/p^\infty$ for some prime $p$. To avoid trouble from the low-dimensional homotopy groups of spheres, we may want to take $p\geq 5$. Note that $\widetilde{H}_*(Y;K)=0$ unless $K$ has characteristic $p$, and that $\widetilde{H}_*(Y;\mathbb{Z}/p)$ is a copy of $\mathbb{Z}/p$ in dimension $3$. Does $Y$ have an $H$-space structure? I do not see an easy way to answer that. Note that $Y\wedge Y$ is a Moore space with $H_5(Y\wedge Y)=\mathbb{Z}/p^\infty$, but that does not immediately give a good hold on $[Y\times Y,Y]$.

Here are some comments about the case where $X$ is not assumed to have finite type. Put $Y=\Sigma X$. For any field $K$, the groups $H_*(Y;K)$ form a Hopf algebra in which all elements of the augmentation ideal are primitive. If $u$ and $v$ lie in the augmentation ideal, then $u$, $v$ and $uv$ are all primitive, which gives $u\otimes v+(-1)^{|u||v|}v\otimes u=0$. If $u$ and $v$ are nonzero, it follows that $|u|=|v|$ and $Ku=Kv$, and $|u|$ is odd unless $K$ has characteristic $2$. Thus, $H_*(Y;K)$ is either $K$ or $K\oplus Ku$ for some element $u$, usually of odd degree. In particular, we see that $Y$ is rationally trivial or an odd-dimensional sphere.

Now consider the groups \begin{align*} A(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}_{(p)}) \\ B(p)_k &= A(p)_k/p \\ C(p)_k &= \widetilde{H}_k(Y;\mathbb{Z}/p) \\ D(p)_k &= \text{ann}(p,A(p)_{k-1}) \end{align*} so that

• $A(p)\otimes\mathbb{Q}$ has dimension $0$ or $1$ over $\mathbb{Q}$
• $C(p)$ has dimension $0$ or $1$ over $\mathbb{Z}/p$
• There is a short exact sequence $B(p)\to C(p)\to D(p)$, so $B(p)$ and $D(p)$ have dimension $0$ or $1$, and at least one of them is zero.

There are a number of different possibilities here.

• If $D(p)=0$ then $A(p)$ is torsion-free and so injects in $A(p)\otimes\mathbb{Q}$. This means that $A(p)=0$ or $A(p)\simeq\mathbb{Z}_{(p)}$ or $A(p)\simeq\mathbb{Q}$.
• If $B(p)=0$ then $A(p)$ is divisible, and so is injective as a $\mathbb{Z}_{(p)}$-module. If we let $T(p)$ denote the torsion part of $A(p)$ then we find that $T(p)$ is also divisible and therefore injective and therefore a summand in $A(p)$. We therefore have $A(p)=T(p)\oplus Q(p)$, where $Q(p)$ is a $\mathbb{Q}$-module. It follows that $A(p)\otimes\mathbb{Q}\simeq Q(p)$, so $Q(p)$ is $0$ or $\mathbb{Q}$. If $D(p)=0$ then $T(p)=0$. If $D(p)=\mathbb{Z}/p$ then I think it follows that $T(p)=\mathbb{Z}/p^\infty$.

The most interesting question arising from this analysis is as follows. Let $Y$ be the Moore space with $H_2(Y)=\mathbb{Z}/p^\infty$ for some prime $p$. To avoid trouble from the low-dimensional homotopy groups of spheres, we may want to take $p\geq 5$. Note that $\widetilde{H}_*(Y;K)=0$ unless $K$ has characteristic $p$, and that $\widetilde{H}_*(Y;\mathbb{Z}/p)$ is a copy of $\mathbb{Z}/p$ in dimension $3$. Does $Y$ have an $H$-space structure? I do not see an easy way to answer that. Note that $Y\wedge Y$ is a Moore space with $H_5(Y\wedge Y)=\mathbb{Z}/p^\infty$, but that does not immediately give a good hold on $[Y\times Y,Y]$.