If $B$ is finitely generated over $k$, then it is certainly finitely generated over $A$, so we have $B=A[x_1,\ldots,x_r]/I$ for some ideal $I$. Furthermore, $C$ becomes an $A$-algebra via the morphism $\varphi$. Now, we get a morphism $\psi_c:A[x_1,\ldots,x_r]\to C$ of $A$-algebras for every $c=(c_1,\ldots,c_r)\in C^r$ by mapping $x_i\mapsto c_i$. If we want a morphism of $A$-algebras $B\to C$, we need to make sure that $I\subseteq\ker(\psi_c)$. Let us assume that $I\ne(1)$, because otherwise we'd have $B=0$, which would be weird.
Let us first assume $A=k$. If $k$ were algebraically closed, we'd know that there is a common zero $a=(a_1,\ldots,a_r)\in Z(I)$ and we could choose $c_i=\varphi(a_i)$. Indeed, it seems that infinite is not enough:
Let $A=C=\mathbb R$ and $B=\mathbb C$. The problem is that you have no algebra morphism $\mathbb C\to \mathbb R$. It would imply that you can map $i$ to some element $\psi(i)\in\mathbb R$ which satisfies $\psi(i)^2=\psi(i^2)=\psi(-1)=-1$.
However, in the case where $k$ is algebraically closed and $A=k$, we can do it. Now, that does not offer much hope: Even if $k$ is algebraically closed, $A$ is not necessarily an algebraically closed extension field.
Edit: Note that we have also established that if such a $\psi$ exists, it does not have to be unique. You might have a lot of choices for your $c$.