I haven't written a complete proof, but I expect your last (and therefore) first question to have a negative answer. Here is what I believe is a counterexample. Let $k$ be the algebraic closure of $\mathbb{Q}(x_1,x_2,\ldots)$ (a countable number of variables). We can view this as a subfield of $\mathbb{C}$. We have a homomorphism $e:K_0(V_k)\to K_0(V_\mathbb{C})$ given by sending the symbol of a variety $[X]$ to $[X\otimes \mathbb{C}]$. This is seen to be surjective because any complex variety can be defined over a finitely generated field, and therefore over $k$. I expect this to be injective as well. It would be enough to show that for (reducible) $k$-varieties $X$ and $Y$, $[X]-[Y]=0$ when it lies in the kernel of $e$. If $e([X]-[Y])=0$, we have finite partitions into locally closed sets $X\otimes \mathbb{C}= \cup X_i$ and $Y\otimes \mathbb{C}= \cup Y_i$ such that $X_i\cong Y_i$. As before, the partitions and isomorphisms should be definable over $k$, so $[X]-[Y]=0$.