These are fairly standard terms, but for the sake of completeness: An ultrafilter $\mathcal{U}$ on $\omega$ is a p-point if whenever $(A_n)_{n<\omega}$ is a partition of $\omega$ such that $A_n \notin \mathcal{U}$ for all $n$, there is an $X \in \mathcal{U}$ such that $X \cap A_n$ is finite for all $n$. $\mathcal{U}$ is Ramsey if the same holds but with $|X\cap A_n| = 1$ for all $n$. Clearly Ramsey ultrafilters are p-point ultrafilters. The name Ramsey comes from the fact that if $\mathcal{U}$ is a Ramsey ultrafilter, then any graph on $\omega$ has a homogeneous set in $\mathcal{U}$.
Fix an edge relation $E$ on $\omega$ making it into a copy of the random graph. (For concreteness, we could say that $nEm$ if and only if the $\min(n,m)$th binary bit of $\max(n,m)$ is $1$.) A remarkable property of the random graph is that for any finite partition $X_0,X_1,\dots,X_{n-1}$ of $\omega$, there is an $i<n$ such that $(X_i,E)$ is isomorphic to $(\omega,E)$. (See Proposition 3 here.) This implies the following.
Proposition. There is an ultrafilter $\mathcal{U}$ on $\omega$ with the property that for any $X \in \mathcal{U}$, there is a $Y \subseteq X$ such that $(Y,E)$ is isomorphic to $(\omega,E)$.
Proof. For each finite partition $P$ of $\omega$, let $F_P$ be the (clopen) set of ultrafilters $\mathcal{U}$ in $\beta \omega \setminus \omega$ satisfying that for the unique $X$ in $\mathcal{U} \cap P$, there is a $Y \subseteq X$ such that $(Y,E)$ is isomorphic to $(\omega,E)$. By the above fact, each $F_P$ is non-empty. The family $\\{F_P:P\text{ a finite partition of }\omega\\}$ has the finite intersection property. To see this, note that if $P_0,P_1,\dots,P_{n-1}$ are finite partitions of $\omega$, then for any mutual refinement $Q$ of the $P_i$'s, $F_Q \subseteq \bigcap_{i<n} F_{P_i}$. Therefore, by compactness, there is an ultrafilter $\mathcal{U}$ such that for every $P$, $\mathcal{U} \in F_P$. For any set $X \in \mathcal{U}$, we have that $\mathcal{U} \in F_{ \\{X,\omega\setminus X\\} }$, so $X$ has the required property. $\square$
Clearly, any such ultrafilter cannot be Ramsey, since an $E$-homogeneous set will contain no subset isomorphic to the random graph. Call such an ultrafilter aggressively non-Ramsey.
Question. Is it consistent with $\mathsf{ZFC}$ that there is an aggressively non-Ramsey p-point?
Best Answer
No, such a filter cannot exist. Suppose $\mathcal U$ is a $p$-point. For $s\in 2^{<\omega}$, let $A_s$ consist of all $m<\omega$ such that $$\forall n\in\mathrm{dom}(s)\ (nEm\Leftrightarrow s(n)=1).$$ Note that there is a unique $x\in 2^\omega$ so that $A_{x\upharpoonright n}\in \mathcal U$ for all $n<\omega$. As $\mathcal U$ is a $p$-point, there is $B\in\mathcal U$ with $B\subseteq^\ast A_{x\upharpoonright n}$ for all $n<\omega$. The graph $(B, E)$ has the following property: For all $n\in B$, all but maybe finitely many $m\in B$ satisfy $$n E m\Leftrightarrow x(n)=1.$$ This easily implies that there is no $C\subseteq B$ so that $(C,E)$ is (isomorphic to) the random graph, so $\mathcal U$ is not aggresively non-Ramsey.