No, it is not true.
Note that the condition for $f(C)$ to have non-zero winding number around $0$ is more-less empty. Indeed, to fix the winding number of $f(C)$ around $0$ one needs to choose infinity. And for more-less any closed curve $\eta$ in $S^2$ that doesn't pass through $0$ one can choose $\infty \in S^2$ so that the winding number of $\eta$ in $S^2\setminus \infty$ around $0$ is non-zero.
Now, for a concrete set of counter-examples suppose that $f: S^3\to S^2$ is ANY map with non-zero differential at a point $x\in S^3$ such that $f(x)\ne 0$. Then take a small ball $U$ containing $x$ such that $0\notin f(U)$ and a curve $\gamma\subset U$ that projects to a small circle in $S^2$. Choose $\infty\in S^2$ such that $f(\gamma)$ separates $0$ from $\infty$. Then clearly the winding number of $f(\gamma)$ around $0$ is $\pm 1$ but $f^{-1}(0)$ doesn't contain a component linked to $c$.
I can see only one way to fix this. Ask $f$ not to be null-homotopic and ask $C$ to be a full premiage of a point $x\in S^2$ different from $0$ ...