No. An almost $P$-space is a topological space where every nonempty $G_{\delta}$-set has a nonempty interior. By the answers to this recent question, there exists compact almost $P$-spaces without any isolated points. However, I claim that if $X$ is a compact $P$-space and $\mu$ is a Borel measure on $X$, then for each nonempty open set $U$ there is a nonempty open subset $V\subseteq U$ with $\mu(V)=0$. Suppose that $U$ is a nonempty open subset of $X$. Then by induction, we can construct a sequence $(U_{n})_{n\in\omega}$ of open sets with $U_{0}\subseteq U,\overline{U_{n+1}}\subseteq U_{n}$ and where $\mu(U_{n+1})<\frac{1}{2}(\mu(U_{n}))$. Let $G=\bigcap_{n}U_{n}$. Then $G$ is a non-empty $G_{\delta}$-set with $\mu(G)=0$. However, since $X$ is an almost $P$-space, there is a nonempty open set $V$ with $V\subseteq G$. Clearly $\mu(V)=0$ but $\emptyset\neq V\subseteq U$.

No. An almost $P$-space is a topological space where every nonempty $G_{\delta}$-set has a nonempty interior. By the answers to this recent question, there exists compact almost $P$-spaces without any isolated points. However, I claim that if $X$ is a compact $P$-space and $\mu$ is a Borel measure on $X$, then for each nonempty open set $U$ there is a nonempty open subset $V\subseteq U$ with $\mu(V)=0$. Suppose that $U$ is a nonempty open subset of $X$. Then by induction, we can construct a sequence $(U_{n})_{n\in\omega}$ of open sets with $U_{0}\subseteq U,\overline{U_{n+1}}\subseteq U_{n}$ and where $\mu(U_{n+1})<\frac{1}{2}(\mu(U_{n}))$. Let $G=\bigcap_{n}U_{n}$. Then $G$ is a non-empty $G_{\delta}$-set with $\mu(G)=0$. However, since $X$ is an almost $P$-space, there is a nonempty open set $V$ with $V\subseteq G$. Clearly $\mu(V)=0$ but $\emptyset\neq V\subseteq U$.