Checking that a torsion-free abelian group has finite rank

(Edited to reflect that $\mathbb Z_{l}$ denotes the $l$-adic integers instead of $\mathbb Z/l\mathbb Z$.)

No. Let $G$ be the additive group of the ring $\mathbb Z[1/p]$, where $p \neq l$ is a prime. Then you can check that $\mathbb Z[1/p] \otimes_{\mathbb Z} \mathbb Z_l = \mathbb Z_l$, essentially because $1/p \in \mathbb Z_l$. This is free of rank one over $\mathbb Z_l$, even though $\mathbb Z[1/p]$ is not finitely-generated as an abelian group.