$\newcommand\ep\epsilon$Getting rid of the instances of $\approx$, one can state the question as follows: 

>Let $f(X):=E(Y|X)$, where $X$ and $Y$ are random variables (r.v.'s) ($X$ possibly a multivariate one) such that $E|Y|<\infty$. Note that $E\ep=0$ and $Cov(f(X),\ep)=0$, where $\ep:=Y-f(X)$. 
>
>Suppose that for a sequence $(g_n)$ of Borel-measurable functions one has $g_n(X)\to f(X)$ in probability (as $n\to\infty$). Does it then follow that 
\begin{equation*}
Cov(g_n(X),\ep_n)\to0, \tag{1}\label{1}	
\end{equation*}
where $\ep_n:=Y-g_n(X)$? 

The answer to this question is: Of course, not. 

Indeed, let e.g. $X$ be a r.v. uniformly distributed on the interval $[-1,1]$, and let $Y:=X$, so that $f(X)=X$ and $\ep=0$. 

Let 
\begin{equation*}
	g_n(X):=X\,1(|X|\ge1/n)+n^2 X\,1(|X|<1/n). 
\end{equation*}
Then $g_n(X)\to X=f(X)$ in probability. 

However, $Eg_n(X)=0$, $\ep_n=Y-g_n(X)=X-g_n(X)=(1-n^2)X\,1(|X|<1/n)$, and hence 
\begin{equation*}
	Cov(g_n(X),\ep_n)=Eg_n(X)\ep_n=n^2(1-n^2)\,EX^2\,1(|X|<1/n) \\ 
	=(1-n^2)/(3n)\to-\infty\ne0. 
\end{equation*}
So, \eqref{1} fails to hold. $\quad\Box$