Update. I've repaired the argument. The idea was to
use the analogue of the usual non-AC arguments, but using a class forcing instead of just Cohen reals.
Theorem. Every model of ZFC has a class forcing extension
that is a model of ZFC, in which there is no class global linear
ordering of the universe that is definable from parameters.
Proof. Let $\mathbb{P}$ be the Easton support class forcing
product $\mathbb{P}=\Pi_{\gamma\text{
reg}}\text{Add}(\gamma,\gamma\cdot 2)$, which forces to add
$\gamma\cdot 2$ many Cohen subsets to every uncountable regular
cardinal $\gamma$ (this is of course isomorphic to adding one).
Suppose that $G\subset\mathbb{P}$ is $V$-generic, and consider the
extension $V[G]$. The standard forcing arguments show that $V[G]$
is a model of ZFC, and in particular, of the axiom of choice.
Meanwhile, I claim that there is no class global linear ordering
of $V[G]$ that is definable in the language of set theory in
$V[G]$ using parameters. Suppose that there were, and that
$\psi(x,y,z)$ is a formula forced by a condition $q\in G$ to
define a linear order when used with the parameter named by $\dot
z$. Let $\gamma$ be a regular cardinal far above the support of
$q$ and any of the conditions appearing in $\dot z$. The forcing
at stage $\gamma$ added the mutually generic sets $g_\alpha$ for
$\alpha\lt\gamma+\gamma$. Let $A$ be the set of all $g_\alpha$
from the first block, that is, for $\alpha\lt\gamma$, and let $B$
be the set of $g_\beta$ from the second block, for
$\gamma\leq\beta\lt\gamma+\gamma$. Fix the corresponding canonical
names $\dot A$ and $\dot B$, as derived from $\dot G$. Suppose
without loss of generality that $A$ precedes $B$ in the definable
linear order, so that $\phi(A,B,z)$ holds in $V[G]$. Fix a
condition $p\in G$ below $q$ such that $p\Vdash\phi(\dot A,\dot
B,\dot z)$. The condition $p$ mentions less than $\gamma$ much
information about the stage $\gamma$ forcing. Thus, it details
fewer than $\gamma$ many bits of fewer than $\gamma$ many sets
each in $A$ and $B$. Furthermore, a density argument shows that
every possible initial segment of a subset of $\gamma$ occurs for
$\gamma$ many of the Cohen sets in each block. So every set in $A$
whose digits are partially specified by $p$ can be matched by a
set in $B$ that agrees with those digits, and vice versa. Since
$p$ specifies fewer than $\gamma$ many bits, there is in $V$ an
automorphism $\pi$ of the stage $\gamma$ forcing that carries out
a permutation of the coordinates, carrying altogether all the
$\alpha$s in the first block to $\beta$s in the second block and
vice versa, in such a way that happens to have $\pi(p)$ and $p$
both in $G$. We may extend $\pi$ to an automorphism of
$\mathbb{P}$, and consider the resulting transformation of names.
Since $\pi$ swaps the two blocks altogether, we have $\dot
A^\pi_G=B$ and $\dot B^\pi_G=A$. The choice of $\gamma$ ensures
that $\dot z^\pi=\dot z$. But since $p\Vdash\phi(\dot A,\dot
B,\dot z)$ we also have $\pi(p)\Vdash\phi(\dot A^\pi,\dot
B^\pi,\dot z^\pi)$, and as $\pi(p)\in G$, this means that
$\phi(B,A,z)$ in $V[G]$, which means that $B$ precedes $A$ in the
linear order, a contradiction. QED
This argument can be used to show, as Ali mentions in his answer, that there can be a definable class of pairs, having no choice function. Specifically, consider all sets of pairs, which would include the pairs of the form $\{A,B\}$ as I denote them in the proof. If we could definably select one or the other, then we fix a condition $p$ forcing which one is selected, and then find an automorphism $\pi$ that swaps the two elements of the pair, while having $\pi(p)$ still in $G$. Thus, the other set must also be selected, a contradiction.
Corollary. Every model of set theory has a class forcing extension with a definable class of unordered pairs, such that no definable class (with parameters) selects exactly one set from each of those pairs.
