$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.
Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let
$$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$
Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.
Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.
Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.
As to the first example:
Let $X$ be a Hausdorff but non-normal space. Let $A, B$ be two closed sets that cannot be separated by disjoint open sets. Define $R$, the equivalence relation as a subset of $X \times X$ by $\Delta \cup (A \times A) \cup (B \times B)$, which is closed, where $\Delta = \{(x,x): x \in X\}$ is the diagonal. So we identify all points in $A$ to a new point $[A]$ and all points in $B$ to a new point $[B]$ in the closure. Let $q$ be the quotient map $X \rightarrow X/R$.
Then if $U \subseteq X/R$ and $V \subseteq X/R$ would be disjoint open neighbourhoods of $[A]$ and $[B]$, then $q^{-1}[A]$ and $q^{-1}[B]$ would be disjoint open sets in $X$, and this cannot be. So $X/R$ is not Hausdorff.
For $X$ we can take the Sorgenfrey plane, or Tychonoff's plank, or $\mathbb{R}^I$ where $I$ is uncountable, to name some concrete spaces that are all Tychonoff (so Hausdorff) but not normal.
I proved directly that if is $X/R$ is Hausdorff, then $R \subseteq X \times X$ is closed in this answer.
A slightly slicker argument: let $\Delta'$ be the diagonal of $(X/R) \times (X/R)$ which is closed as $X/R$ is Hausdorff, and note that $q \times q: X \times X \rightarrow (X/R) \times (X/R)$ is continuous and $R = (q \times q)^{-1}[\Delta']$ is thus closed.
So closedness of the relation is necessary but not sufficient for Hausdorff $X$.
As to the second question: suppose that $f: X \rightarrow Y$ is a quotient map onto the Hausdorff space $Y$. Define $R_f = \{(x,x') \in X^2 : f(x) = f(x')\}$, the equivalence relation induced by $f$. Standard facts on quotient spaces show that $X/R_f$ is homeomorphic to $Y$ using the map incuced by $f$. Let $q: X \rightarrow X/R_f$ be the quotient map then it's easy to see that $q^{-1}[q[A]] = f^{-1}[f[A]]$ for all $A \subseteq X$. It follows that $q$ is closed iff $f$ is closed. And as $R_f = (f \times f)^{-1}[\Delta_Y]$, where $\Delta_Y$ is the (closed as $Y$ is Hausdorff) diagonal of $Y$, $R_f$ is a closed relation.
So e.g. if $f$ is open, continuous (so quotient), but not closed, like the projection from $\mathbb{R}^2$ onto one of its factors, then $R_f$ is an example of a closed relation with Hausdorff (even metric) quotient, such that $q$ is not a closed map.
As to the final wondering: if $C \subset X$ and $x$ would be in $\overline{C}\setminus C$, $(x,x)$ would be in $\overline{C \times C}\setminus (C \times C)$. So a non-closed set has a non-closed square.
Best Answer
After working on this for a bit, I think came up with a solution. I'm sharing here for everyone.
I want to show that $(0,0)$ and $(0,1)$ in $X$ could not be separated. Let $[0,0]$ be the equivalence class of $(0,0)$ - similar notation will be used for $(0,1)$ - and let $U,V$ two open sets in $X$ s.t. $$[0,0] \in U \\ [0,1] \in V$$ Since $\pi$ is continuous $\pi^{-1}(U),\pi^{-1}(V)$ are open in $Q$ and contains $(0,0),(0,1)$. Since we are in $\mathbb{R}^2$ we have two open balls one centered in $(0,0)$ and one in $(0,1)$. Let $\epsilon$ be the minimum radius, If $(x,y)\in B_\epsilon(0,0)$ and $x>0$ this implies that $$\sqrt{x^2+ (1-(1-y))^2} = \sqrt{x^2+y^2} <\epsilon$$ so $(x,1-y) \in B_\epsilon(0,1)$ but $$U \supset \pi\big(B_\epsilon(0,0)\big)\ni \pi(x,y) =[x,y]=[x,1-y]=\pi(x,1-y)\in\pi (B_\epsilon(0,1))\subset V $$ So we can conclude that $U\cap V\neq \emptyset $.