No, this fails not only for selective ultrafilters but for all non-principal ultrafilters $\mathcal F$ on  $\omega$. 

The main ingredient in the proof is the theorem that, if an ultrafilter $\mathcal U$ on a set $X$ is sent to itself by some map $g:X\to X$ (meaning that $\mathcal U=g(\mathcal U):=\{A\subseteq X:g^{-1}(A)\in\mathcal U\}$), then the set of points in $X$ fixed by $g$ has to be in $\mathcal U$. This easily implies that, if $\phi:X\to Y$ is a bijection and $g:X\to Y$ is an arbitrary map and $\mathcal U$ is an ultrafilter on $X$ with $\phi(\mathcal U)=g(\mathcal U)$, then $\{x\in X:\phi(x)=g(x)\}\in\mathcal U$. 

In your situation, suppose toward a contradiction that $\phi$ were a bijection as in the question. Let $\mathcal G$ be the filter on $[\omega]^2$ generated by the sets $[B]^2$ for $B\in\mathcal F$. ($\mathcal G$ may or may not be an ultrafilter, depending on whether or not $\mathcal F$ is selective.) Your requirement on $\phi$ implies that each set in $\mathcal G$ meets each set of the form $\phi^{-1}(A)$ for $A\in\mathcal F$.  Since the intersection of two sets of the form $[B]^2$ for $B\in\mathcal F$ is again a set of that form, and since the intersection of two sets of the form $\phi^{-1}(A)$ for $A\in\mathcal F$ is again a set of that form, it follows that the sets of these two forms generate a proper filter. Let $\mathcal U$ be an ultrafilter on $[\omega]^2$ extending that filter. So we have $[B]^2\in\mathcal U$ and $\phi^{-1}(A)\in\mathcal U$ for all $A,B\in\mathcal F$.

It follows that $\phi(\mathcal U)=\mathcal F$. It also follows that $g(\mathcal U)=\mathcal F$, where $g:[\omega]^2\to\omega$ is the function sending each pair $\{x<y\}\in[\omega]^2$ to its smaller member $x$. By the theorem cited above, we have that the set $E=\{\{x<y\}\in[\omega]^2:x=\phi(\{x,y\})\}$ is in $\mathcal U$. Since $\phi$ is one-to-one, $E$ contains, for any $x\in\omega$, at most one pair whose smaller element is $x$. Thus, there is a function $f:\omega\to\omega$ such that $E\subseteq\{\{x,f(x)\}:x\in\omega$. Note that $x$ here is the smaller element (the one given by $g$) in the pair $\{x,f(x)\}$, and so we have  $x<f(x)$ for all $x$.

Consider any two sets $B,C\in\mathcal F$. Since $[B\cap C]^2$ and $E$ are both in $\mathcal U$, they have a nonempty intersection. In particular, there is some $x\in C$ with $f(x)\in B$, i.e., with $x\in f^{-1}(B)$. I've just shown that $f^{-1}(B)$ intersects every set $C\in\mathcal F$; since $\mathcal F$ is an ultrafilter, it follows that $f^{-1}(B)\in\mathcal F$. And since this holds for all $B\in\mathcal F$, we have $f(\mathcal F)=\mathcal F$. By the theorem cited at the beginning, $f(x)=x$ for $\mathcal F$-almost all $x$. This contradicts the fact that $x<f(x)$ for all $x$.