I need to change the notation a little to make it a little more consistent with stats references and easier to follow. I'll switch to the orginal poster's notation at the end.
$$X \sim binomial(n, p)$$
$$f(x,n,p) = {n \choose p} p^x (1-p)^{n-x}$$
$$F(x,n,p) = P(X \le x) = \sum_{k=0}^{x} f(k,n,p)$$
For the truncated distribution, you can find many references to show this:
$$f(x,n,p|a < X \le b) = \frac{g(x,n,p)}{F(b) - F(a)}$$
where $g(x,n,p) = f(x,n,p)$ if $a < x \le b$ and $g(x,n,p) = 0$ otherwise.
$$E(X | a < X \le b) = \sum_{a+1}^b xf(x,n,p|a<X\le b) \\
= \frac{1}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b x {n \choose p} p^x (1-p)^{n-x} \\
= \frac{np}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b \frac{(n-1)!}{(n-x)!(x-1)!} p^{x-1} (1-p)^{n-x} \\
= \frac{np}{F(b,n,p)-F(a,n,p)} \sum_{a}^{b-1} \frac{(m)!}{(m-y)!(y)!} p^{y} (1-p)^{m-y} \ \ where\ \ y=x-1\ \ and\ \ m=n-1 \\
= \frac{np}{F(b,n,p)-F(a,n,p)} [F(b-1,n-1,p) - F(a-1,n-1,p)]$$
$$E(X(X-1) | a<X\le b) = \sum_{a+1}^b x(x-1)f(x,n,p|a<X\le b) \\
= \frac{1}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b x(x-1) {n \choose p} p^x (1-p)^{n-x} \\
= \frac{n(n-1)p^2}{F(b,n,p)-F(a,n,p)} \sum_{a+1}^b \frac{(n-2)!}{(n-x)!(x-2)!} p^{x-2} (1-p)^{n-x} \\
= \frac{n(n-1)p^2}{F(b,n,p)-F(a,n,p)} \sum_{a-1}^{b-2} \frac{(m)!}{(m-y)!(y)!} p^{y} (1-p)^{m-y} \ \ where\ \ y=x-2\ \ and\ \ m=n-2 \\
= \frac{n(n-1)p^2}{F(b,n,p)-F(a,n,p)} [F(b-2,n-2,p) - F(a-2,n-2,p)]$$
$$V(X | a<X\le b) = E(X^2| a<X\le b) - E(X| a<X\le b)^2 \\
= E(X(X-1)| a<X\le b) + E(X| a<X\le b) - E(X| a<X\le b)^2$$
$$V(X | a<X\le b) = \frac{n(n-1)p^2[F(b-2,n-2,p) - F(a-2,n-2,p)]}{F(b,n,p)-F(a,n,p)} + \frac{np[F(b-1,n-1,p) - F(a-1,n-1,p)]}{F(b,n,p)-F(a,n,p)} - \left[ \frac{np[F(b-1,n-1,p) - F(a-1,n-1,p)]}{F(b,n,p)-F(a,n,p)} \right]^2$$
Now, switching to the original notation and situation, I can get the ratio of variances as a function of $q$, but not quite the nice inequality you were looking for.
$$\frac{V(R|q-1<R\le n)}{V(R)} = \frac{1}{np(1-p)} \left( \frac{n(n-1)p^2[1 - F(q-3,n-2,p)]}{1-F(q-1,n,p)} + \frac{np[1 - F(q-2,n-1,p)]}{1-F(q-1,n,p)} - \left[ \frac{np[1 - F(q-2,n-1,p)]}{1-F(q-1,n,p)} \right]^2 \right)$$
The next step would be to show by a series arguments and operations to the right hand side, that
$$\frac{V(R|q-1<R\le n)}{V(R)} \le \frac{n-\frac{np(1-F(q-2,n-1,p))}{(1-F(q-1,n,p))}}{n-np} = \frac{N-ER^*}{N-ER}$$
Update on the Normal Approximation
Matt F's post using the normal approximation inspired me to continue, but I'm not sure that path reaches the ultimate goal of proving the inequality under all conditions.
See Wikipedia's Truncated Normal Distribution page for these facts
Let $\alpha = (a-\mu)/\sigma$ and $\beta = (b-\mu)/\sigma$ and $\phi$ is the standard normal PDF and $\Phi$ is the normal CDF.
$$E(X|a<X<b) = \mu - \sigma \frac{\phi(\beta)-\phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}$$
$$V(X|a<X<b) = \sigma^2 \left[ 1-\frac{\beta \phi(\beta)-\alpha \phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)} - \left(\frac{\phi(\beta)- \phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)}\right)^2\right]$$
Now, substituting in for this specific situation: with $W = 1 - \Phi(\alpha)$ and $\alpha = (q-np)/\sqrt{np(1-p)}$
$$\frac{V(R|q<R\le n)}{V(R)} = \frac{\sigma^2 \left[1 + \frac{\alpha \phi(\alpha)}{W} - \frac{\phi(\alpha)^2}{W^2}\right]}{\sigma^2} \le \frac{n-ER^*}{n-ER} = \frac{n-\left( np + \frac{\sqrt{np(1-p)} \phi(\alpha)}{W} \right)}{n - np}$$
$$ 1+\frac{(q-np)\phi(\alpha)}{W \sqrt{np(1-p)}} - \frac{\phi(\alpha)^2}{W^2} \le 1 - \frac{\phi(\alpha)\sqrt{np(1-p)}}{(n-np)W} $$
cancelling terms
$$\frac{q-np}{\sqrt{n(1-p)}} - \frac{\sqrt{p}\phi(\alpha)}{W} \le -\frac{p}{\sqrt{n(1-p)}}$$
To guarantee the inequality of interest:
$$q \le (n-1)p + \frac{\sqrt{np(1-p)}\ \ \phi\left( \frac{(q-np)}{\sqrt{np(1-p)}}\right) }{1-\Phi\left( \frac{(q-np)}{\sqrt{np(1-p)}} \right)}$$