Let $X$ be a set equipped with a well-ordering $\leq$ as well as an (unrelated) equivalence relation $E$.
We define a relation $\leq'$ on $X/E$, given for $a,b \in X/E$ by $a \leq' b$ iff there is some $x \in a$ such that $x \leq y$ for all $y \in b$.
Claim: $\leq'$ is a total order on $X/E$.
Proof:
Note $\leq'$ is reflexive as $X$ is well-ordered; and transitive; and antisymmetric since $X/E$ is a partition, of $X$.
Finally we show any two elements of $X/E$ are $\leq'$-related. Aiming for a contradiction, suppose not; suppose some $a,b \in X/E$ are not $\leq'$-related. It follows that for every $x \in a$ there is some $y \in b$ which is less than $x$; and similarly for every $y \in b$ there is some $x \in a$ which is less than $y$. ("Less than" meaning $\leq$; $\neq$ also because $a,b$ are disjoint.) But this cannot happen, since we can use well-orderedness to pick a $\leq$-least element $z$ of $a \cup b$, which must lie in one or the other, and now whichever of $a,b$ doesn't contain $z$ is now unable to have an element less than $z$. Thus we have a contradiction.
My question is whether this total order $\leq'$ can fail to be a well-ordering on $X/E$?
My guess is yes, but I'm not really sure how to come up with a counterexample. Would anyone have any suggestions on how to proceed? (Also, is the reasoning in the above Claim valid?)
Update: actually I think it should be a well-ordering along the following reasoning:
Suppose $\leq'$ is not a well-order. Then we can find some collection $A \subset X/E$ having no $\leq'$-element; but if we select $x$ a $\leq$-element of $\bigcup A$ then the unique $a \in A$ for which $x \in A$ should be $\leq'$-least in $A$?
You're actually overcomplicating things by thinking in terms of quotients in the first place - in fact, your construction is better thought of as a suborder of the original order.
Every $E$-class ${\bf e}$ has a least element since $\mathcal{X}=(X,\le)$ is a well-ordering. This least element determines the whole position of the set ${\bf e}$ in your order: given $E$-classes ${\bf e},{\bf f}$ with least elements $e,f$ respectively, we have $${\bf e}\le'{\bf f}\quad\iff\quad e\le f.$$ Letting $Y$ be the set of $\le$-least elements of $E$-classes, we have that your proposed ordering is isomorphic to the suborder of $\mathcal{X}$ with domain $Y$. Since a suborder of a well-order is again a well-order, the answer to your question is yes.
Note that this trick crucially relies on the fact that the original order was a well-order. In the presence of a mere linear order, your construction just gives a partial order: consider as our starting order the integers $\mathbb{Z}$ with the usual order, and let $E$ partition $\mathbb{Z}$ into the even and odd numbers.
Perhaps more transparently, if $\mathcal{X}$ is a linear order and $E$ is an equivalence relation on $X$ such that every $E$-class has an $\le$-least element, then $\mathcal{X}/E$ (defined as in the OP) is canonically isomorphic to a suborder of $\mathcal{X}$. The point is then that if $\mathcal{X}$ is a well-order, then every $E$ has this additional property.