On the functoriality of scalar extensions of local rings (edited) Note. I have edited my question to make it more transparent, following some very good comments that I received. I am sorry if it is a bit long.
A local homomorphism of local rings $(A,\mathfrak{m})\stackrel{\varphi}{\longrightarrow}(B,\mathfrak{n})$ is called a scalar extension (terminology due to Hans Schoutens) if: 


*

*$\varphi(\mathfrak{m})B=\mathfrak{n}$, and

*$\varphi$ is a flat extension.


Let's fix a field $K$ (algebraically closed, if you wish) and let $\mathscr{C}_K$ be the category of Noetherian local rings whose residue field is a subfield of $K$, with morphisms being local homomorphisms. 
Question A. Is there a functorial way of producing scalar extensions with a prescribed residue field? More precisely, is it possible to define a functor $F_K:\mathscr{C}_K\rightarrow\mathscr{C}_K$ in such a way that for every $A\in\mathscr{C}_K$ the local ring $F_K(A)$ is a scalar extension of $A$ with residue field $K$?
Here are some things that I know about this question:
(1) Grothendieck proved that scalar extensions with prescribed residue field always exist:
Theorem. (EGA III, Proposition 10.3.1, page 20). Let $(A,\mathfrak{m})$ be Noetherian local ring with residue field $k$, and let $K$ be a field extension of $k$. Then there exists a scalar extension $(A,\mathfrak{m})\stackrel{\varphi}{\longrightarrow}(B,\mathfrak{n})$ from $A$ to a Noetherian local ring $B$, with the property that $B/\mathfrak{n}$ is $k$-isomorphic to $K$. 
Grothendieck's construction of the desired scalar extension depends on various 'choices' that he makes in his proof, and hence, does not produce a unique answer. For this reason I think it is hopeless to get a functor there.
(2) Various mathematicians have used a different method to construct scalar extensions with prescribed residue field, which seems 'more hopeful' to be functorial. In [b] (pp. 776-777) Kunz calls a special case of this construction the constant field extension. A version of this construction in the equicharacteristic case appears in [a] (pp. 18-19, 6.3). A more detailed description of this method can be found in [c] (pp. 4-7) and in [d] (pp. 36-38). I describe it in the equicharacteristic case: Given a local ring $(A,\mathfrak{m},k)$ and a field extension $K$ of $k$, take a coefficient field $k\hookrightarrow\hat{A}$ and complete $\hat{A}\otimes_kK$ with respect to the ideal $\mathfrak{m}(\hat{A}\otimes_kK)$. This is your $F_K(A)$. It is easy to see that this $F_K(A)$ is an scalar extension of $A$ with residue field $K$. ($F_K(A)$ depends on the choice of a coefficient field of $\hat{A}$, but is unique up to isomorphism).
Question B. Is the $F_K(\:\cdot\:)$ that was just described a functor from $\mathscr{C}_K$ to $\mathscr{C}_K$? To clarify the question, if $A_1\stackrel{\psi}{\longrightarrow} A_2$ is a local homomorphism of Noetherian local rings in $\mathscr{C}_K$, then does $\psi$ extend to a local homomorphism $B_1:=F_K(A_1)\rightarrow B_2:=F_K(A_2)$?
I can see how the method described in [c] provides an affirmative answer in equicharacteristic $0$ to Question B (it comes down to the fact that in equicharacteristic $0$ every maximal subfield of a complete local ring is a coefficient field) but I don't see how the method of [c] would still work in equicharacteristic $p>0$. I haven't checked the mixed characteristic case, yet, because I thought the equicharacteristic case is easier and if it cannot be settled positively, then there is even less hope for the mixed characteristic.
References.
a. M. Hochster and C. Huneke, $F$-regularity, test elements, and smooth base change, Trans. Amer. Math. Soc., 346 (1994).
b. E. Kunz, Characterizations of regular local rings of characteristic $p$, Amer. Jour. of Math., 41 (1969).
c. H. Schoutens, Classifying singularities up to analytic extensions of scalars, Ann. of Pure and Applied Logic, 162, (2011) (also available on the Arxiv, here).
d. H. Schoutens, The use of ultraproducts in commutative algebra, Lecture Notes in Mathematics, 1999, Speringer (2010).
 A: 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.
