I've thought about this and discussed this question with someone else. In fact it suffices to prove that any complete intersection of two quadrics containing a curve of odd degree must contain a line - hence the answer to your second question is affirmative. I haven't checked all details yet but I might edit this later.

By a theorem of Amer, an intersection of quadrics $f = g = 0$ contains a linear space of dimension $r$ over $k$ iff the quadric given by $f + tg = 0$ contains a linear space of dimension $r$ over $k(t)$. Hence it suffices to prove that any quadric which contains a curve of odd degree actually contains a line.

To do this, note that such a quadric must have a rational point: if you cut by a generic hyperplane, you will get a finite set of points, at least one of which has odd degree. But Springer's theorem says that if a quadric contains a point in an odd degree extension of the base field, then it must have a rational point. So the quadric is isotropic and we can split off a hyperbolic plane. With the "rest" of the quadric you can repeat the argument, although you probably have to be careful with configuration issues - I'd have to write down this part more carefully. Hence you will get two hyperbolic planes, hence a line.