Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$:
$$E\min(X,n-X)=n-2E|X-n/2|.$$
You can upper bound $E|X-n/2|$ using Jensen's inequality: $E|X-n/2|\le\sqrt{E(X-n/2)^2}$.
The latter, if I'm not mistaken, evaluates to $$ n\sqrt{ p(1-p)/n+p^2-p+1/4 } =:nF(p).$$
For $n$ sufficiently large and $p$ sufficiently small (certainly, $p\le 0.65$; the exact value can be easily computed), we have $2F(p)\le c(1-p)$ for some universal $c>0$. That means that $n-2F(p)n\ge cnp$, so your conjecture holds for this range of $p$. I'm confident that with a bit more care you can extend the result to all $p$, perhaps with a worse constant.