Show that $a \otimes 1$ will vanish in $A\otimes_{\Bbb{Z}}\Bbb{Q}$ only when $a=0$. 
Let $A$ be a $\mathbb{Z}$-module. I want to see that if $a \otimes 1=0$ in $A\otimes_{\Bbb{Z}}\Bbb{Q}$, then $a=0$. I'll assume $A$ is torsion free.

Can I use that there exists an injective homomorphism from $A \otimes \mathbb{Q}$ to $Hom(A^{*} \times \mathbb{Q}^{*}, \mathbb{Z})$ to show that $a=0$? The homomorphism is given by $f(a \otimes 1)(v^{*},w^{*})=v^{*}(a)w^{*}(1)$
Every $\alpha: \mathbb{K} \to \mathbb{Z}$ verifies $\alpha(1) \alpha\left(\frac{1}{\alpha(1)}\right)=\alpha(1)$ so $\alpha(1)$ is $1$ or $-1$. Then $f(a \otimes 1)(v^{*},w^{*})=\pm v^{*}(a)$ showing that $a$ must be zero.   
 A: Assuming that you have seen the fact that $\Bbb{Q}$ is a flat $\Bbb{Z}$-module you can argue as follows.
Because $A$ is assumed to be torsion free, the submodule generated by the element $a$ is isomorphic to $\Bbb{Z}$. So $f:\Bbb{Z}\to A, n\mapsto na$ is an injective homomorphism. Flatness means that
$$
f\otimes id: \Bbb{Z}\otimes\Bbb{Q}\to A\otimes\Bbb{Q}, n\otimes q\mapsto na\otimes q
$$
is also injective.
But here $\Bbb{Z}\otimes\Bbb{Q}\cong\Bbb{Q}$, so $1\otimes 1\neq0$. By injectivity $f(1\otimes 1)=a\otimes1\neq0$, too.
A: As I commented, your approach is doomed because there are no homomorphisms from $\mathbb{Q}$ to $\mathbb{Z}$.  Here are a couple different possible approaches you could take.
Approach 1: Consider the localization $A_{(0)}$ of the $\mathbb{Z}$-module $A$ obtained by inverting every nonzero element of $\mathbb{Z}$.  Explicitly, $A_{(0)}$ consists of equivalence classes of fractions $a/n$ where $a\in A$ and $n\in\mathbb{Z}\setminus\{0\}$, where $a/n=b/m$ iff there exists a nonzero $k\in\mathbb{Z}$ such that $kma=knb$.  Construct a bilinear map $A\times \mathbb{Q}\to A_{(0)}$, and prove that it has the universal property of the tensor product, so in fact $A_{(0)}\cong A\otimes \mathbb{Q}$.  Now use the explicit description of $A_{(0)}$ in terms of fractions and the fact that $A$ is torsion-free to show that $a\otimes 1=0$ implies $a=0$.
Approach 2:  First, prove that the statement you want is true if $A$ is a free $\mathbb{Z}$-module.  Now note that if $a\otimes 1=0$ in $A\otimes\mathbb{Q}$, there is a finitely generated subgroup $A_0\subseteq A$ such that $a\in A_0$ and $a\otimes 1=0$ in $A_0\otimes\mathbb{Q}$ (to prove this, think about the explicit construction of the tensor product as a quotient of the free $\mathbb{Z}$-module on formal expressions $a\otimes b$).  Now use the classification of finitely generated abelian groups to show $A_0$ is free (this is the step where you need $A$ to be torsion-free).
