$\newcommand{\R}{\mathbb{R}}
\newcommand{\vpi}{\varphi}$
The answer is: the condition 

>$E(f(X,Y)|Y)=\vpi(Y)$ for all $f$ such that $f(X,Y)$ has a finite expectation 

holds iff $X$ and $Y$ are independent. 

Indeed, if $X$ and $Y$ are independent, then for each Borel subset $B$ of $\R$
\begin{align*}
	EI\{Y\in B\}\vpi(Y)&=\int P(Y\in dy)I\{y\in B\}\vpi(y) \\ 
	&=\int P(Y\in dy)I\{y\in B\}Ef(X,y) \\ 
		&=\int P(Y\in dy)I\{y\in B\}\int P(X\in dx)f(x,y) \\ 
		&=\int P(X\in dx,Y\in dy)I\{y\in B\}f(x,y) \\ 
		&=EI\{Y\in B\}f(X,Y),
\end{align*}
where $I$ is the indicator. So, $E(f(X,Y)|Y)=\vpi(Y)$. 

Vice versa, suppose that $E(f(X,Y)|Y)=\vpi(Y)$ for all $f$ such that $f(X,Y)$ has a finite expectation. Take any Borel subsets $A$ and $B$ of $\R$, and let $f(x,y):=I\{x\in A\}I\{y\in B\}$ for all real $x,y$. Then for each $y\in\R$
\begin{equation}
\vpi(y)=Ef(X,y)=EI\{X\in A\}I\{y\in B\}=P(X\in A)I\{y\in B\}	
\end{equation}
and 
\begin{align*}
P(X\in A)P(Y\in B)&=EP(X\in A)I\{Y\in B\} \\ 
&=EI\{Y\in B\}\vpi(Y) \\ 
	&=EI\{Y\in B\}f(X,Y) \\
	&=EI\{X\in A\}I\{Y\in B\}=P(X\in A,Y\in B),
\end{align*}
so that $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$.