Regarding $\mathbb{C}$ as a set, i.e. considering the set $\mathbb{R}^2$, can any total order be defined on this set that has the least upper bound property?
Of course it is know that if any such order exists it will not be "compatible" with the field operations on $\mathbb{C}$ as in the definition of an order field.
On
Since $\Bbb R$ and $\Bbb C$ have the same cardinality, you can find a bijection between them and simply transport the structure of $\Bbb R$ to $\Bbb C$.
One can be a bit more clever about it, and notice that $(\Bbb R,+)$ and $(\Bbb C,+)$ are isomorphic as $\Bbb Q$-vector spaces, so one can require that the bijection above preserves addition too.
Yes. Take a bijection $b\colon\mathbb{C}\longrightarrow\mathbb{R}$ and define$$z\preccurlyeq w\iff f(z)\leqslant f(w).$$