Warning: I know nothing about this subject, but found the question interesting so decided to learn something about it. Approach the following with caution.

Consider the ring of integers $\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}] \subset \mathbb{Q}[\sqrt{-15}]$. This has class number 2, and the ideal
$$I:=(2, \tfrac{1+\sqrt{-15}}{2})$$
generates the ideal class group: in other words this is a projective module of rank 1, and is not free.

Writing $\zeta = \tfrac{1+\sqrt{-15}}{2}$ we have
$$\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}] = \mathbb{Z}[\zeta]/(\zeta^2-\zeta+4),$$
and so its 2-adic completion is
$$(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2 = \mathbb{Z}^\hat{}_2[\zeta]/(\zeta^2-\zeta+4).$$

Using [http://www.numbertheory.org/php/2adic.html] we find that $-15$ has a 2-adic square root $\eta$ starting $(1,0,0,1,1,0,1,1,0,0,...)$. Thus $1+\eta = (0,1,0,1,1,0,1,1,0,0,...)$ is divisible by 2, and we set $\bar{\eta} = \tfrac{1+\eta}{2}=(1,0,1,1,0,1,1,0,0,...)$. Hence we have
$$\zeta^2-\zeta+4 = (\zeta - \bar{\eta})(\zeta -\tfrac{4}{\bar{\eta}}) \in \mathbb{Z}^\hat{}_2[{\zeta}],$$
and so
$$\mathbb{Z}^\hat{}_2[\zeta]/(\zeta^2-\zeta+4) = \mathbb{Z}^\hat{}_2[\zeta]/((\zeta - \bar{\eta})(\zeta -\tfrac{4}{\bar{\eta}})) \cong \mathbb{Z}^\hat{}_2[\zeta]/(\zeta - \bar{\eta}) \times \mathbb{Z}^\hat{}_2[\zeta]/(\zeta -\tfrac{4}{\bar{\eta}})$$
which is isomorphic to $\mathbb{Z}^\hat{}_2 \times \mathbb{Z}^\hat{}_2$ as a ring. As $K_0(\mathbb{Z}^\hat{}_2 \times \mathbb{Z}^\hat{}_2) = K_0(\mathbb{Z}^\hat{}_2) \oplus K_0(\mathbb{Z}^\hat{}_2) = \mathbb{Z} \oplus \mathbb{Z}$ is torsion-free, the conclusion is that
$$K_0(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}]) \to K_0((\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2)$$
fails to be injective: in fact, its kernel is precisely $\mathbb{Z}/2$.

The relevance of this to the question is that, as I understand it, 2-adic $TC$ is insensitive to 2-adic completion of the ring (at least for rings which are finitely generated as abelian groups), that is,
$$TC(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2 \to TC((\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2)^\hat{}_2$$
is an equivalence. It follows from the evident commutative square that
$$K_0(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2 \to TC_0(\mathbb{Z}[\tfrac{1+\sqrt{-15}}{2}])^\hat{}_2$$
is not injective.