My comment, with more details: No, $E = H\mathbb{Q}$ and $F = \Sigma\mathbb{Z}$ gives a counterexample. $H_0(E) = \mathbb{Q}$ and $H_i(E) = 0$ for $i \neq 0$, so the universal coefficient theorem gives $[E,F] = H^1(E) = \mathrm{Ext}^1_\mathbb{Z}(\mathbb{Q},\mathbb{Z})$, which is an uncountable $\mathbb{Q}$ vector space. In contrast, $[E,F_\mathbb{Q}] = \mathrm{Ext}^1_\mathbb{Z}(\mathbb{Q},\mathbb{Q}) = 0$, so $j_*$ is not injective in this case.
I don't think it suffices that $E$ and $F$ are bounded below and have finitely generated homotopy groups, either. For instance, take $E = \Sigma^\infty_+ BG$ for a finite group $G$ and take $F = ku$ the connective topological $K$-theory spectrum. Then $[E,F]$ is the completed representation ring of $G$, by the Atiyah--Segal completion theorem. For $G = \mathbb{Z}/2\mathbb{Z}$ it is additively isomorphic to $\mathbb{Z} \oplus \mathbb{Z}_2$, so it rationalizes to $\mathbb{Q} \oplus \mathbb{Q}_2$. In contrast, $$[E,F_\mathbb{Q}] = \prod_i H^{2i}(BG;\mathbb{Q}) = \mathbb{Q}.$$