Welcome to MathOverflow! The answer to your question is yes. Indeed, let $X$ and $Y$ be independent random elements of the set $\{0,1\}^n$ whose distributions are percolation measures, which latter thus satisfy the [Harris--FKG inequality, Proposition 2.3][1]: $Ef(X)g(X)\ge Ef(X)\, Eg(X)$ and $Ef(Y)g(Y)\ge Ef(Y)\, Eg(Y)$ for all bounded nondecreasing functions $f$ and $g$. We have to show that for any such function $f$ and $g$ we have $Ef(X\vee Y)g(X\vee Y)\ge Ef(X\vee Y)\, Eg(X\vee Y)$, where $X\vee Y$ is coordinate-wise maximum of $X$ and $Y$. 

Letting $F(y):=Ef(X\vee y)$ and $G(y):=Eg(X\vee y)$ for $y\in\{0,1\}^n$, we have 
\begin{align}
Ef(X\vee Y)g(X\vee Y)&=\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)g(X\vee y) \\
&\ge\sum_{y\in\{0,1\}^n} P(Y=y)Ef(X\vee y)\,Eg(X\vee y) \\
&=\sum_{y\in\{0,1\}^n} P(Y=y)F(y)\,G(y) \\ 
&=EF(Y)\,G(Y) \\ 
&\ge EF(Y)\,EG(Y) \\ 
&= Ef(X\vee Y)\, Eg(X\vee Y),
\end{align}
as desired. the first inequality in the above multiline display is an instance of the Harris--FKG inequality for $X$ and nondecreasing functions $x\mapsto f(x\vee y)$ and $x\mapsto g(x\vee y)$, and the second inequality there is an instance of the Harris--FKG inequality for $Y$ and nondecreasing functions $F$ and $G$.  


  [1]: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&ved=2ahUKEwjUnoSox9zjAhXYHM0KHfC0C1YQFjABegQIBBAB&url=http%3A%2F%2Fwww.ihes.fr%2F~duminil%2Fpubli%2F2017percolation.pdf&usg=AOvVaw2VxEQnQ7slBZi7kxwwstBd