Here is a counter-example in the case when $G$ is 2-generated. Let $G=<a>\times <b>$ where $a$ has finite order $p>1$ and $b$ has infinite order. Let $H=<c>$, where $c=ab^p$. Then the infinite cyclic group $H$ is not contained in a free factor of $G$. Indeed, otherwise, $c$ admits $p$-th root: $c=x^p$. Then $x=a^n b$, but $x^p=(a^{pn} b^p)=b^p$. Contradiction.
Edit: Here are details for Will's comments:
Suppose that $G=T\times R$, where $T$ is the torsion subgroup of $G$. Let $F\subset G$ be a torsion-free subgroup.
Assume now that $F\subset \tilde{F}$, where $\tilde{F}$ is a
direct torsion-free factor of $G$. Then it is easily seen that $G=T\times \tilde{F} \times L$. Thus, $G/F\cong T\times (\tilde{F}/F) \times L$ and the sequence
$$
0\to T\to G/F\to G/<F,T>\to 0
$$
clearly splits. Conversely, suppose that the above sequence splits. Then $G/F=T\times R$. Taking preimage of $1\times R$ under the homomorphism
$G\to G/F\to T\times R$, we obtain a torsion-free subgroup $\tilde{F}\subset G$ which contains $F$. By construction $T$ maps isomorphically to $G/\tilde{F}$. Thus, $G=T\times \tilde{F}$. qed
