Section of the homology functor on spectra

I think that the answer to your question is no, but not for the reason you may suspect. Let us try to do the "stupid" case first: fix an abelian group $A$ and you want a spectrum $X$ such that $H_0(X) =A$ and $H_i(X)=0$ for $i\neq 0$. Such a spectrum is called Moore spectrum and it always exists. However it is not functorial in $A$, that is every map between abelian groups may be realized as a map between the corresponding Moore spectra but not uniquely so. However you can realize Moore spectra functorially if you consider only 2-divisible groups.

Finally, this case is in fact the only problem you have. Supposing you were able to realize Moore spectra functorially, you could realize graded abelian groups easily enough by taking wedges of suspensions.

I thought I will post the only reference I have found about non-existence of Moore space functor as an answer. It is due to Carlsson, "A counterexample to a conjecture of Steenrod" https://link.springer.com/article/10.1007%2FBF01393939 and there is a nice discussion on nLab https://ncatlab.org/nlab/show/Moore+space#nonfunctoriality_of_the_construction