Is the given binomial sum almost everywhere negative as $K\to\infty$?

Letting $p=p(\theta)=1-\theta$ and $q=q(\theta)={1-2\theta\over 1-\theta},$ you want to show that $$\sum_{i=0}^{K/2} {K\choose i} p^i (1-p)^{K-i}<{1\over 2}\sum_{i=0}^{K/2} {K\choose i} q^i (1-q)^{K-i},\tag1$$ for sufficiently large $K$.

Trivial case: For $1/3\leq \theta <1/2$, the sum on the left converges to zero and the sum on the right converges to a positive number, so the inequality $(1)$ is true for large $K$.

Remaining case: Suppose $0< \theta <1/3$. We prove the inequality of the sums working term by term. It suffices to show that $$p^i(1-p)^{K-i}<{1\over 2}q^i(1-q)^{K-i}\tag2$$ for all $0\leq i\leq K/2$, when $K$ is large enough.

Note that ${p(1-q)\over q(1-p)}={1-\theta\over 1-2\theta}> 1$ and ${p(1-p)\over q(1-q)}={(1-\theta)^3\over 1-2\theta}< 1$ (for $0<\theta<1/3$). Therefore, $$\left({1-p\over 1-q}\right)^K\left(p(1-q)\over q(1-p)\right)^i \leq \left({1-p\over 1-q}\right)^K\left(p(1-q)\over q(1-p)\right)^{K/2} = \left({p(1-p)\over q(1-q)}\right)^{K/2}.\tag3$$

The right hand side of $(3)$ can be made less than $1/2$ by taking $K$ sufficiently large, giving the inequality (2) and hence the inequality (1).