The answer is **negative**: If $f'(x) = 0$ "too often", then $R_f$ may fail to be equal to one almost everywhere.

Let $C$ be a fat Cantor set, let $I_n = (a_n, b_n)$ ($n \geqslant 2$) be the sequence of all finite components of the complement of $C$, and let $f$ be a differentiable function with the following properties:

$f(x) = 0$ for $x \in C$;

on $I_n$, $f$ is a smooth bump of a fixed shape, supported in $\tfrac1n I_n$ (the middle $n$th part of $I_n$), and with maximum equal to $|I_n|^2$ (and hence the integral of $f$ over $I$ is equal to $\tfrac1n |I_n|^3$).

Then $f'(x) = 0$ for $x \in C$ by the first property, so $f$ is everywhere differentiable. On the other hand, it is rather straigthforward to see that there is a constant $C$ such that for each $n$ and $t \in I_n = (a_n, b_n)$ we have
$$ \int_{a_n}^t f(y) dy \leqslant \frac{C}{n} (t - a_n) \sup_{y \in [a_n, t]} f(y) $$
and
$$ \int_t^{b_n} f(y) dy \leqslant \frac{C}{n} (b_n - t) \sup_{y \in [t, b_n]} f(y) . $$
It follows that if $x \in C$ and $r$ is small enough, so that $B_r(x)$ is disjoint with $\tfrac12 I_2 \cup \tfrac13 I_3 \cup \ldots \cup \tfrac1{n-1} I_{n-1}$, then
$$ \int_{B_r(x)} f(y) dy \leqslant \frac{C}{n} |B_r(x)| \sup_{y \in B_r(x)} f(y) . $$
This, in turn, implies that $R_f(x) = \infty$ for every $x \in C$.

Thus, $R_f(x) \ne 1$ on a set of positive Lebesgue measure.

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