Yes. Let $\mathscr{U}$ be an ultrafilter on $X$, and let $a\in X$ be arbitrary. Since $\mathscr{U}$ is an ultrafilter, either $\{a\}\in\mathscr{U}$, or $X\setminus\{a\}\in\mathscr{U}$. If $\{a\}\in\mathscr{U}$, then $\mathscr{F}_a\subseteq\mathscr{U}$, and therefore $\mathscr{U}=\mathscr{F}_a$. If $X\setminus\{a\}\in\mathscr{U}$ for each $a\in X$, then $X\setminus F\in\mathscr{U}$ for each finite $F\subseteq X$; this is easily proved by induction on $|F|$.
Alternatively, you can argue as follows. Let $\mathscr{F}$ be the cofinite filter on $X$. If $\mathscr{U}\supseteq\mathscr{F}$, then $$\bigcap\mathscr{U}\subseteq\bigcap\mathscr{F}=\varnothing\;,$$ so $\mathscr{U}$ cannot be a principal filter: $\bigcap\mathscr{F}_a=\{a\}$ for each $a\in X$. If, on the other hand, $\mathscr{U}\nsupseteq\mathscr{F}$, then there is a finite $F\subseteq X$ such that $X\setminus F\notin\mathscr{U}$. Let $F=\{x_1,\ldots,x_n\}$; clearly $$F=\{x_1\}\cup\{x_2\}\cup\ldots\cup\{x_n\}\;,$$ a finite union, so there is exactly one $k\in\{1,\dots,n\}$ such that $\{x_k\}\in\mathscr{U}$, and it follows immediately that $\mathscr{U}=\mathscr{F}_{x_k}$, the principal ultrafilter over $x_k$.