For any two functions $f,g$ the sets $\{n\in\omega:f(n)\leq g(n)\}$ and $\{n\in\omega:g(n)\leq f(n)\}$ cover $\omega$, so one of them must be in $\mathcal U$. Hence we have a dichotomy $f\leq_{\mathcal U}g$ or $f\leq_{\mathcal U}g$.

It follows that every unbounded family is dominating: if $B$ is an unbounded with respect to $\leq_{\mathcal U}$family, then for any $f\in\omega^\omega$ there is $g\in B$ such that $f\not\geq_{\mathcal U}g$ and hence $f\leq_{\mathcal U}g$, hence $B$ is dominating. Since every dominating set is clearly unbounded, it follows that $\frak b_{\mathcal U}=\frak d_{\mathcal U}$. However, it is consistent that $\frak b<\frak d$, so it is consistent that $\frak b_{\mathcal U}\neq\frak b$ or $\frak d_{\mathcal U}\neq\frak d$.

We can say something more: every set unbounded with respect to $\leq_{\mathcal U}$ is unbounded with respect to $\leq^*$ (because $f\not\leq_{\mathcal U}g\Rightarrow f\not\leq^*g$) and every set dominating with respect to $\leq^*$ is dominating with respect to $\leq_{\mathcal U}$ (because $f\leq^*g\Rightarrow f\leq_{\mathcal U}g$), so $\frak b\leq\frak b_{\mathcal U}=\frak d_{\mathcal U}\leq\frak d$, hence it's consistent that one of these inequalities is strict.

Of course assuming $CH$ we would have $\frak b=\frak b_{\mathcal U}=\frak d_{\mathcal U}=\frak d=\frak c$, hence the only open question left is whether we can (consistently) have $\frak b<\frak b_{\mathcal U}=\frak d_{\mathcal U}<\frak d$, which is a question I can't answer.