My sense is that this statement is probably false even though I don't immediately see a counter example. The statement is easily checked to be true for very stable bundles, so you when you look for a counter example you will need to look for a wobbly (i.e. stable but not very stable) bundle.
Suppose that $E$ is a rank two bundle with $E$ stable and $\det E = \mathcal{O}$. We are trying to decide if it can happen that for all traceless $\theta : E \to E\otimes K$ it follows that $\det (\theta)$ has a double zero. This condition looks quadratic but it is in fact linear. Given a point $x \in C$, the traceless condition tells you that $\det (\theta)$ will vanish at $x$ if and only if $\theta(x)$ is a nilpotent $K_{x}$-valued endomorphism of $E_{x}$. Furthermore $\det (\theta)$ will vanish to second order at $x$ if and only if $\theta_{x}$ is the zero endomorphism. So your question is: can we find a wobbly bundle $E$ so that for every Higgs field $\theta$ on $E$ there exists a point $x \in C$ so that $\theta(x) = 0$? Since the bundle $\text{End}_{0}(E)\otimes K$ is of rank $3$, for every point $x$ this is a linear condition of codimension $\leq 3$.
Notice however that for a stable $E$, the natural map $$ \text{End}_{0}(E_{x})\otimes K_{x} \longrightarrow H^{1}(C,\text{End}_{0}(E)\otimes K(-x))) $$ is surjective.
Therefore the question is: what are the possible dimensions of $H^{1}(C,\text{End}_{0}(E)\otimes K(-x)))$ or dually of $H^{0}(C,\text{End}_{0}(E)(x))$ as $x$ varies in $C$? Concretely $E$ will give you a counter example if and only if for all $x$ you have $h^{0}(C,\text{End}_{0}(E)(x)) \geq 2$. So you have to analyse all wobbly bundles $E$ for this property. Since the wobbly bundles are explicitly classified this will be straightforward. If the property holds for some wobbly bundle you will find your counter example. If it fails for all of them, this will give you the proof you were looking for.