Here is an interesting situation to think about, which might produce a counterexample:

Take two copies of $\mathbb{P}^2$ and consider the standard quadratic (Cremona) birational transformation $\tau\colon (x:y:z) \mapsto (u:v:w)=(yz:xz:xy)$ which restricts to an isomorphism between the open sets $\{xyz\neq 0\}$ of one $\mathbb{P}^2$ and $\{uvw\neq 0\}$ of the other; now call $X$ the scheme obtained by gluing the two $\mathbb{P}^2$ together by identifying these two open sets by means of $\tau$. In other words, $X$ is the projective plane with a triangle doubled, but doubled in two different ways.

If we take the closure of the diagonal $\Delta \subseteq X\times X$, it contains every pair where the first part lies on the line $x=0$ (of the first $\mathbb{P}^2$), say, and the second is the point $v=w=0$ (of the second $\mathbb{P}^2$); this closure is not an equivalence relation, but the equivalence relation it generates identifies all the points of the lines $x=0$, $y=0$, $z=0$ and $u=0$, $v=0$ and $w=0$.

So if we try to "separify" $X$ by somehow quotienting by the closure of the diagonal, we are led to contract the three lines $x=0$, $y=0$ and $z=0$ in $\mathbb{P}^2$ to a single point. Now I believe this cannot be done. This suggests that there is no universal morphism from $X$ to a separated scheme.

Here is an interesting situation to think about, which might produce a counterexample:

Take two copies of $\mathbb{P}^2$ and consider the standard quadratic (Cremona) birational transformation $\tau\colon (x:y:z) \mapsto (u:v:w)=(yz:xz:xy)$ which restricts to an isomorphism between the open sets $\{xyz\neq 0\}$ of one $\mathbb{P}^2$ and $\{uvw\neq 0\}$ of the other; now call $X$ the scheme obtained by gluing the two $\mathbb{P}^2$ together by identifying these two open sets by means of $\tau$. In other words, $X$ is the projective plane with a triangle doubled, but doubled in two different ways.

If we take the closure of the diagonal $\Delta \subseteq X\times X$, it contains every pair where the first part lies on the line $x=0$ (of the first $\mathbb{P}^2$), say, and the second is the point $v=w=0$ (of the second $\mathbb{P}^2$); this closure is not an equivalence relation, but the equivalence relation it generates identifies all the points of the lines $x=0$, $y=0$, $z=0$ and $u=0$, $v=0$ and $w=0$.

So if we try to "separify" $X$ by somehow quotienting by the closure of the diagonal, we are led to contract the three lines $x=0$, $y=0$ and $z=0$ in $\mathbb{P}^2$ to a single point. Now I believe this cannot be done (except by contracting everything to a point). This suggests that there is no universal morphism from $X$ to a separated scheme (or else that this separification should be a point, which is suspicious).