It is straightforward to show that there is an ultrafilter $\mathcal{U}_0$ on the positive integers such that every element $A\in \mathcal{U}_0$ satisfies $$d^\ast(A):=\limsup_{n\to +\infty} \frac{\lvert\{a\in A : a \leq n\}\rvert}{n}>0. $$
Fix $\epsilon\in (0,\frac{1}{2}]$. Is there any hope that there is an ultrafilter $\mathcal{U}_\epsilon$ on the positive integers such that every element $A\in \mathcal{U}_\epsilon$ satisfies $$d^\ast(A)>\epsilon? $$
If $\mathcal{U}$ is an ultrafilter on $\mathbb{N}$ and $\mathbb{N}=A_1\sqcup\ldots\sqcup A_k$, then exactly one $A_i$ is in $\mathcal{U}$ (exercise).
As a consequence of this fact, the answer to your question is no. For any $\epsilon>0$ we can write $\mathbb{N}$ as the disjoint union of finitely many sets, each of which has asymptotic density $<\epsilon$. For instance, look at the sets $A_i=\{n: \exists m(n=km+i)\}$ for large enough $k$.