I post this as an answer since it is too long, actually answers Question B and sheds some light on Question A. The example is taken from Eisenbud, Commutative Algebra, Exercise 7.17b.
Let $A_1=\mathbf F_p(t)$, $A_2=\mathbf F_p(u)[[x]]$, $\psi\colon A_1\to A_2$, $t\mapsto u^p+x$. On the residue fields, $\psi$ induces an isomorphism $\mathbf F_p(t)\cong\mathbf F_p(u^p)$. If $K$ is any extension field of $\mathbf F_p(u)$, the $F_K$ in Question B has $F_K(A_1)=K$ and $F_K(A_2)=K[[x]]$ with the obvious homomorphisms $A\to F_K(A)$. However, the diagram
$$\begin{array}{ccc} \mathbf F_p(t) & \xrightarrow{\psi} & \mathbf F_p(u)[[x]] \\ \downarrow && \downarrow \\
K && K[[x]]\end{array}$$
cannot be completed since $t$ becomes the $p$-th power $u^p$ in $K$, whereas it is mapped to $u^p+x$ in $K[[x]]$ which is not a $p$-th power.
As for Question A, depending on the exact interpretation, any $F_K$ should satisfy $F_K(k)=K$ for subfields $k\subset K$, so this example shows that $k[[x]]\to F_K(k[[x]])$ cannot just be the canonical homomorphism $k[[x]]\to K[[x]]$ in general.
EDIT: The above assumes, contrary to what you said in the comments, that the embedding of the residue field into $K$ is part of the data in $\mathscr C_K$. Otherwise, you can still apply the same argument if you assume $K$ algebraically closed (so that the image of $t$ is still a $p$-th power). However, I think that in this case even the simpler problem of choosing a natural embedding into $K$ for all fields in $\mathscr C_K$ is already impossible except in trivial cases.
