Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$:

$$E\min(X,n-X)=n/2-2E|X-n/2|.$$

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
$n/2-2F(p)n\ge cnp$, 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. Update: this "confidence" has proven misplaced, see Dmitry Krachun's answer below!]