First, a remark: the free part of $A(K)$ is a quotient, not a sub, and so it is possible that a point of infinite order in $A(K)$ could have non-trivial image in $\Phi_A(X)$. Probably what you mean is that $\mathcal A^0(X)$ and $A(K)$ have the same free rank.
Regarding torsion, my interpretation of your question is that you are asking about the map $A(K)_{tors} \to \Phi_A(X)$, and are curious is to whether or not it can have a kernel (so that some part of $A(K)_{tors}$ is contained in $\mathcal A^0(X)$).
As far as I know, this varies a lot depending on the particular abelian variety, but in particular cases it has been quite intensively studied. For example, if $p$ is prime and $A =J_0(p)$ is the Jacobian of the modular curve $X_0(p)$, then Mazur showed in his Eisenstein ideal paper showed that the map $A(\mathbb Q)_{tors} \to \Phi_A( \mathrm{Spec} \\, \mathbb Z)$ is an isomorphism. I generalized this to subabelian varities $A$ of $J_0(p)$ in my paper here.
For an example in some sense opposite to this, see this paper of Conrad, Edixhoven, and Stein, in which they show that if $A = J_1(p)$ (again $p$ is prime), then $\Phi_A(\mathrm{Spec} \\, \mathbb Z)$ is trivial, so that $\mathcal A^0 = \mathcal A$.