Here are some ideas, building on Deanne Yang's excellent idea to use the support function. Let our hypothetical unit circle be centered at the origin. I'll also imagine that the origin is in $R$, maybe someone else will see how to remove this.
Any curve with a flex has radius of curvature $\infty$ at the flex, so $\partial R$ has no flexes and thus $R$ is convex. (This follows from a theorem of Tietze that I learned about here.) Thus, any ray from the origin meets $\partial R$ at a unique point $r(\theta) (\cos \theta, \sin \theta)$. Letting $h$ be the sucpport function Deanne introduces, note that we have $h(\theta) \geq r(\theta)$.
Let's suppose we have $2m$ points $\theta_1$, $\theta_2$, ..., $\theta_{2m}$, such that $r(\theta)$ is $>1$ on $(\theta_{2j-1}, \theta_{2j})$ and $<1$ on $(\theta_{2j},\theta_{2j+1})$. Since $h \geq r$, we also have $h \geq 1$ on $(\theta_{2j-1}, \theta_{2j})$. On the other hand, choose $\phi \in (\theta_{2j},\theta_{2j+1})$. Then, since $R$ is convex, there must be some $\theta$ such that the supporting hyperplane $(\cos \theta) x + (\sin \theta) y = h(\theta)$ separates $(\cos \phi, \sin \phi)$ from $R$. For this $\theta$, we must have $h(\theta) < (\cos \theta)(\cos \phi) + (\sin \theta) (\sin \phi) = \cos (\theta-\phi) \leq 1$, and drawing a picture shows that we must have $\theta \in (\theta_{2j}, \theta_{2j+1})$. So, in each $(\theta_{2j}, \theta_{2j+1})$, there is some place where $h$ is $<1$.
We are thus reduced to the question: If $h$ is periodic modulo $2 \pi$ and $h+h'' < 1$, can $h$ cross the value $1$ more than twice in a period? One may as well put $f = h-1$, so that the equation is $f+f''<0$, and ask about zeroes of $f$ instead.
One idea I had about how to approach this was to show that, if $f+f''<0$ then two zeroes of $f$ couldn't be that close together. If $f$ is negative between the zeroes, then I can show that they are at least $2$ apart. Proof: Suppose for the sake of contradiction that $f(0) = f(a) = 0$ for $0<a<2$ with $f+f''<0$ and $f$ negative on $(0,a)$. Let $f$ be minimized at $b$ and, WLOG, rescale $f$ so that $f(b) = -1$. By the mean value theorem, there are $0 < c < b < d < 0$ with $f'(c) = -1/b$ and $f'(d) = 1/(a-b)$ so, by the mean value theorem again, there is $e \in (b,d)$ with $f''(e) = \tfrac{1/(a-b) + 1/b}{d-c} = \tfrac{a}{b(a-b)(d-c)} \geq \tfrac{a}{(a/2)^2 a} = 4/a^2$. But, also, $f(e) \geq -1$. So $4/a^2-1 < 0$ and $a \geq 2$. $\square$ I have no idea how close this bound is to tight, except that the sine function shows that we can't enlarge $2$ all the way up to $\pi$.
I couldn't find a comparable bound if $f$ is positive between the zeroes. Note that the function $2-x^2$ demonstrates that the zeroes can be as close together as $2 \sqrt{2} < \pi$, demonstrating that the sine functional is not optimal as one might guess.