Your attempt is almost correct.
Let the demand function be $D(p) = p^{-\epsilon}$ and the cost function some $C(D(p))$ so that $R(p) = D(p)p = p^{1-\epsilon}$, right?
Then, I'd rewrite your first and second order conditions - as you only provided the expressions but it would be great to write them in a form of equations (or inequalities). So that:
First order condition: $R'(p) - C'(D(p))D'(p) =0$
And the second order: $R''(p) - C''\left(D(p)\right)\cdot D'(p)^2 - C'(D(p))D''(p) < 0$
This is the moment when we kind of split our approaches. I also do agree that the second term in second order condition is negative. However, it is not necessarily $R''(p)$ that is negative - rather the whole relation $$R''(p) - C'(D(p))D''(p)<0 \;(*)$$ Now, we want the second order condition to hold for every $p$ that satisfies the first order condition. Thus, I would substitute the $C'(D(p))$ from the first condition to the $(*)$ and hence, I end up with: $$R''(p) - \frac{R'(p)D''(p)}{D'(p)} <0 \;(**)$$
Now it's time for the brute force. Calculate all derivatives in $(**)$.
$$ -\epsilon(1-\epsilon)p^{-1-\epsilon} - \frac{(1-\epsilon)p^{-\epsilon} \cdot -\epsilon(-1-\epsilon)p^{-2-\epsilon}}{-\epsilon p^{-1-\epsilon}} <0 \implies \epsilon > 1$$ for both, $p$ and $\epsilon$ positive.
P.S. If you think about it, letting $\epsilon<1$ would imply that $R'(p)>0$ and thus, the first order condition would have no solutions. Why?
P.P.S. I believe a similar exercise is in Industrial Organization: Markets and Strategies by Belleflamme and Peitz in their chapter about "monopoly pricing strategies" but with the inverse demands $P(q)$.