I had to think for a while to understand Scott's answer (or at least, what I suspect he meant by his answer), and in the end there were enough details to sort out that I thought they were worth posting. It ended up being too long to post as a comment, so here it is as a separate answer. Unless it's all nonsense, of course....

Let {$x_{\alpha}$} be a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, and let $L$ be the intermediate field that they generate, so that $\mathbb{C}$ is the algebraic closure of $L$ in $\mathbb{C}$. Take also a collection of open disks $D_{\alpha}$ in $\mathbb{C}$ such that any collection of points $y_{\alpha} \in D_{\alpha}$ is dense in $\mathbb{C}$ in the usual topology. Now for each $\alpha$, take $x_\alpha$ and multiply it by an appropriate root of unity and a rational number so that the result $y_\alpha$ lies in $D_\alpha$. The collection {$y_{\alpha}$} is still algebraically independent over $\mathbb{Q}$, because a dependence gives an algebraic dependence of {$x_\alpha$} over some finite extension of $\mathbb{Q}$, which implies the existence of an algebraic dependence over $\mathbb{Q}$ as well.

So there exists $\sigma : L \to \mathbb{C}$ sending $x_{\alpha} \mapsto y_{\alpha}$. Now by the usual fact that field embeddings into algebraically closed fields can be extended across algebraic extensions, $\sigma$ extends to a map $\mathbb{C} \to \mathbb{C}$. But note that by construction $\sigma$ is surjective! The image contains each $y_\alpha$, and it contains all the roots of unity, so it contains all the $x_\alpha$'s; thus the image is an algebraic closure of $L$ in $\mathbb{C}$, hence all of $\mathbb{C}$.

In particular $\mathbb{C}$ is a quadratic extension of $\sigma(\mathbb{R})$, obtained by adjoining $\sigma(i)$. But finally $\sigma(\mathbb{R})$ is dense in $\mathbb{C}$ since its image contains all the $y_\alpha$'s, and so giving $\sigma(\mathbb{R})$ the norm induced from the usual norm on $\mathbb{C}$, we get a normed field $\sigma(\mathbb{R})$ whose completion is exactly $\mathbb{C}$, i.e., a quadratic extension of it. Thus the answer to your second question is no as well.

I had to think for a while to understand Scott's answer (or at least, what I suspect he meant by his answer), and in the end there were enough details to sort out that I thought they were worth posting. It ended up being too long to post as a comment, so here it is as a separate answer. Unless it's all nonsense, of course....

Let {$x_{\alpha}$} be a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$, and let $L$ be the intermediate field that they generate, so that $\mathbb{C}$ is the algebraic closure of $L$ in $\mathbb{C}$. Take also a collection of open disks $D_{\alpha}$ in $\mathbb{C}$ such that any collection of points $y_{\alpha} \in D_{\alpha}$ is dense in $\mathbb{C}$ in the usual topology. Now for each $\alpha$, take $x_\alpha$ and multiply it by an appropriate root of unity and a rational number so that the result $y_\alpha$ lies in $D_\alpha$. The collection {$y_{\alpha}$} is still algebraically independent over $\mathbb{Q}$, because a dependence gives an algebraic dependence of {$x_\alpha$} over some finite extension of $\mathbb{Q}$, which implies the existence of an algebraic dependence over $\mathbb{Q}$ as well.

So there exists $\sigma : L \to \mathbb{C}$ sending $x_{\alpha} \mapsto y_{\alpha}$. Now by the usual fact that field embeddings into algebraically closed fields can be extended across algebraic extensions, $\sigma$ extends to a map $\mathbb{C} \to \mathbb{C}$. But note that by construction $\sigma$ is surjective! The image contains each $y_\alpha$, and it contains all the roots of unity, so it contains all the $x_\alpha$'s; thus the image is an algebraic closure of $L$ in $\mathbb{C}$, hence all of $\mathbb{C}$.

In particular $\mathbb{C}$ is a quadratic extension of $\sigma(\mathbb{R})$, obtained by adjoining $\sigma(i)$. But finally $\sigma(\mathbb{R})$ is dense in $\mathbb{C}$ since its image contains all the $y_\alpha$'s, and so giving $\sigma(\mathbb{R})$ the norm induced from the usual norm on $\mathbb{C}$, we get a normed field $\sigma(\mathbb{R})$ whose completion is exactly $\mathbb{C}$, i.e., a quadratic extension of it. Thus the answer to your second question actually yes.