If we don't require continuity at $a$, the answer is an easy "no". The answer is still "no" if we require continuity, but not quite as easily.
If we look at the function $g\colon (0,1) \to \mathbb{R}$,
$$g(x) = (-1)^n\quad\text{ if } \frac{1}{(n+1)!} \leqslant x < \frac{1}{n!},$$
and consider
$$f(x) = \int_0^x g(t)\,dt,$$
we have an almost-example: For $a_k = \frac{1}{k!}$ we have
\begin{align}
\frac{f(a_{2k})-f(0)}{a_{2k}} &= (2k)!\int_0^{1/(2k)!} g(t)\,dt\\
&= (2k)!\biggl(\frac{1}{(2k)!} - \frac{1}{(2k+1)!}\biggr) + (2k)!\int_0^{1/(2k+1)!} g(t)\,dt\\
&= \frac{2k}{2k+1} + O\biggl(\frac{1}{2k+1}\biggr)
\end{align}
and
\begin{align}
\frac{f(a_{2k+1}) - f(0)}{a_{2k+1}} &= (2k+1)!\int_0^{1/(2k+1)!}g(t)\,dt\\
&= -(2k+1)!\biggl(\frac{1}{(2k+1)!}-\frac{1}{(2k+2)!}\biggr) + (2k+1)!\int_0^{1/(2k+2)!} g(t)\,dt\\
&= -\frac{2k+1}{2k+2} + O\biggl(\frac{1}{2k+2}\biggr),
\end{align}
so the difference quotients $\frac{f(x)-f(0)}{x}$ vary from very close to $1$ to very close to $-1$ and back when $x$ traverses the interval $[a_{2k+2},a_{2k}]$, and the set of all limit points of the difference quotients for sequences $x_k \searrow 0$ is the interval $[-1,1]$.
The $f$ considered above is only an almost-example, since $f$ is not differentiable on the whole interval $(0,1)$, there are countably many points where $f$ isn't differentiable.
We can convert it to a true example if we alter $g$ to linearly interpolate instead of jump at the points $a_k$ if we choose a steep enough slope for the linear interpolation. Then the modification $\tilde{g}$ is continuous, and its integral $\tilde{f}$ is differentiable on $(0,1)$ but not right-differentiable at $0$.
