This is a pretty long question that I have been struggling with for a while, any tips or suggestions would be awesome!
Let $A, B$ be two C*-algebras, the later algebra is represented on a complex Hilbert space $H$, $a, b\in A^+$ two positive elements of norm less than or equal to 1 and $p: A\to B$ a (completely if needed) positive linear map. Functional calculus gives us that $$(a-b)^2\geq 0 \quad\text{ and } \quad (a-b)^2 \leq [(a-b)^2]^{1/2}$$ where the later inequality holds since $||a - b|| \leq 1$ so the square root is monotone on the spectrum of $(a - b)^2$. We hence get
$$\langle p( (a-b)^2)x, x\rangle \leq \langle p( [(a-b)^{2}]^{1/2})x, x\rangle \quad \forall x\in H$$where $[(a-b)^2]^{1/2}$ denotes the unique positive square root of $(a-b)^2$
My question is the following - is it true in general that if $c$ is an arbitrary root of $(a-b)^2$ (like $(a-b)$) do we have $$\langle p( (a-b)^2)x, x\rangle \leq |\langle p(c)x, x\rangle |$$
No, even if $p$ is ucp and $c$ is selfadjoint.
Take $A=B=M_2(\mathbb C)$, $p(a)=\operatorname{Tr}(a)\,I$, and $a=I$, $b=0$, $$ c=\begin{bmatrix} t&s\\ s&-t\end{bmatrix} $$ with $s^2+t^2=1$. Then $p(c)=0$, so your inequality only holds for $x=0$.