Does the order of "unbounded" quantifiers matter?

No. For example, let $P(x,y)$ mean $x\ne y$, and assume there are at least two things in the universe we quantify over.

Then $\exists x \forall y (x\ne y)$ is false, but $\forall y \exists x (x \ne y)$ is true.

Inspired by Henning's answer...

Let a and b be two distinct objects in the domain of quantification $U$ (the "universe" set). We have:

$$a\in U$$ $$b\in U$$ $$a\neq b$$

Yes, I know the OP said "unbounded quantifiers," but I think you really need to make the "universe" set explicit to understand the principle at work here.

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Part 1

Required to prove: $\neg \exists x\in U: \forall y\in U:x\neq y$

Suppose to the contrary. Let $c$ be such that $c\in U$ and $\forall y\in U: c\neq y$. This leads to the obvious contradiction $c\neq c$.

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Part 2

Required to prove: $\forall y\in U: \exists x\in U: x\neq y$

Let $c$ be such that $c\in U$. Consider two cases.

Case 1: Suppose $a=c$.

Then $b\neq c$ by substitution and symmetry. Since $b\in U$, we then have $\exists x\in U: x\neq c$.

Case 2: Suppose $a\neq c$.

Since $a\in U$, we also have $\exists x\in U: x\neq c$.

In both cases, we have $\exists x\in U: x\neq c$. Generalizing on $c$, we have, as required:

$\forall y\in U: \exists x\in U: x\neq y$