The answer to your first question is 'No'.
Let $g$ be the function $S^1\to S^1$ which starts at $1$, moves anticlockwise to $-1$, then moves clockwise thrice $1 + \sqrt{2}$ times as fast once round the circle back to $-1$, and then moves anticlockwise back to $1$ again. This function $g$ has degree $0$ and $\hat{g}(0)=0$. The function $f(\theta) = g(\theta) e^{i\theta}$, which moves at a constant speed, therefore has degree $1$ and $\hat{f}(1)=0$.
You may reasonably complain at this point that $f$ is not differentiable and certainly not simple, but $f$ can be deformed very slightly so that it bounds a topological disc and makes smooth turns.
