For arbitrary (selective) ultrafilter $\mathcal{F}$ does there exist bijection $\phi:[\omega]^2\to\omega$ with the property: $\forall B\in\mathcal{F} : \phi([B]^2)\in\mathcal{F}$ ?
No, this fails not only for selective ultrafilters but for all nonprincipal 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 onetoone, $E$ 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 $x$.

$\begingroup$ Thank you for your answer. Where can I read about the theorem you mention in beginning? $\endgroup$ – ar.grig Feb 27 at 20:24

2$\begingroup$ The theorem used in the solution is due to Katetov (1967), you can find the short paper where it appeared through the link: dml.cz/bitstream/handle/10338.dmlcz/105124/…. You can also find Solovay's rendition of Katetov's theorem through the link: math.berkeley.edu/~solovay/Preprints/Rudin_Keisler.pdf $\endgroup$ – Ali Enayat Feb 28 at 0:14

$\begingroup$ @AliEnayat: Thank you. I got the links. But what about reference of Solovay's rendition which could be cited in article ? $\endgroup$ – ar.grig Mar 2 at 9:16

$\begingroup$ @ar.grig To my knowledge, Solovay's rendition has never appeared in print; his argument is "essentially the same" as Katetov's. As pointed out in the following source, Katetov's theorem can be viewed as a special case of the BruijnErdős theorem: dml.cz/bitstream/handle/10338.dmlcz/126182/… $\endgroup$ – Ali Enayat Mar 3 at 17:44