Edit (May 2017):
I initially claimed to answer the question, but I precisely only obtain the following:
(a) There is no countable torsion-free reduced abelian group $A$ containing copies of all $\mathbf{Q}$-rank 2 torsion-free countable reduced abelian groups.
Fix a prime $p$. In a group $G$, let us define a $p$-maximal subgroup as the kernel of a nontrivial homomorphism onto the quasi-cyclic group $C_{p^\infty}=\mathbf{Z}[1/p]/\mathbf{Z}$.
Claim 1: in $\mathbf{Q}^2$, there are uncountably (continuum) many $p$-maximal subgroup, only countably many of which are non-reduced.
Proof of claim 1: first we count the non-reduced ones. Each non-reduced one is in the $\mathrm{GL}_2(\mathbf{Q})$-orbit of one containing $\mathbf{Q}\times\{0\}$. Then we conclude by arguing that $\mathrm{Hom}(\mathbf{Q},C_{p^\infty})$ is countable (exercise). It remains to check there are uncountably many $p$-maximal subgroups, and actually uncountably containing $\mathbf{Z}_{(p)}^2$ ($\mathbf{Z}_{(p)}$ is the set of rationals with denominator coprime to $p$). We have $\mathbf{Q}/\mathbf{Z}_{(p)}\simeq C_{p^\infty}$, and $C_{p^\infty}^2$ has uncountably (continuum) many subgroups (with quotient isomorphic to $C_{p^\infty}$ as well). This finishes the proof of the first claim. (Reference for these standard counting facts: B. Boyer, Enumeration theorems in infinite Abelian groups, Proc. AMS 7:565–570, 1956, who first characterized abelian groups with countably many subgroups.
Claim 2: in $\mathbf{Q}^2$, there are uncountably (continuum) many non-isomorphic reduced $p$-maximal subgroups.
Proof: by Claim 1 there are uncountably many. Since any isomorphism between any two extends (uniquely) to a automorphism of $\mathbf{Q}^2$, and since the automorphism group of $\mathbf{Q}^2$ is countable (this is $\mathrm{GL}_2(\mathbf{Q})$), we deduce Claim 2.
Let us finish the proof of (a). Assume by contradiction that $A$ exists. For a subgroup $H$ of $A$; denote by $\widehat{H}$ the inverse image in $A$ of the torsion subgroup of $A/H$. Note that $\widehat{H}$ and $H$ have the same $\mathbf{Q}$-rank, and more precisely that if $\overline{H}$ is the divisible hull of $H$, there is a canonical embedding of $\widehat{H}$ into $\overline{H}$ that is the identity on $H$; it cannot be surjective because $A$ is reduced.
Now assume that $H$ is $p$-maximal in its divisible hull. Then $\overline{H}/H\simeq C_{p^\infty}$, so by nonsurjectivity, we deduce that $\widehat{H}/H$ is finite.
Furthermore, if $H$ has finite $\mathbf{Q}$-rank $d$, $\widehat{H}$ is equal to $\widehat{H}$ for any free abelian subgroup of rank $d$; in particular, since $A$ is countable, there are only countably possibilities for $\widehat{H}$.
If we assume that $H$ is both $p$-maximal in its divisible hull and has finite $\mathbf{Q}$-rank, we combine the previous two paragraphs; since $\widehat{H}$ has only countably many subgroups of finite index (finitely many for each given index), we deduce that there are countably many possibilities for $H$.
We thus conclude that $A$ contains copies of only countably many of the $p$-maximal subgroups of $\mathbf{Q}^2$. Given Claim 2, this means that $A$ fails to contain some reduced $p$-maximal subgroup of $\mathbf{Q}^2$. This concludes the proof of (a).
Remark: the above argument also shows, regardless of the continuum hypothesis, the following:
The least cardinal of a reduced torsion-free abelian group containing isomorphic copies of all reduced subgroups of $\mathbf{Q}^2$ is $2^{\aleph_0}$.
Edit (May 2017): I've removed a confusing paragraph I initially wrote, which does not address the intended question anyway.
I think I made a confusion between being reduced and residually finite for abelian groups. For abelian groups, clearly residually finite implies reduced. The converse is true for torsion-free abelian groups. But it is easy to find counterexamples among countable torsion abelian $p$-groups.
If I could remove the first occurrence of "torsion-free" in (a), it would answer the question. But the quotient of even a residually finite abelian group by its torsion subgroup can fail to be reduced (a classical instance being $\prod_p\mathbf{Z}/p\mathbf{Z}$, or the preimage of a copy of $\mathbf{Q}$ modulo the torsion if we want a countable example). So we cannot naively reduce to the torsion-free case.
