The paper linked formulates quadratic variation in a measure-theoretic framework. The references therein may also be of interest. As a disclaimer, I did not read this paper very closely, nor is this a research area I am familiar with. I imagine being able to access measure-theoretic tools might offer some interesting approaches to solving this problem.
In the paper, the authors assign positive finite (Hausdorff) measures to functions of bounded quadratic variation. Their Theorem 20 proves that for every positive finite measure on $[0,1]$, there is a function of finite quadratic variation that generates this measure (and I believe has quadratic variation related to the variation norm of the measure. So a nonzero finite measure would correspond with a nonzero but finite quadratic variation function).
The paper: https://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05209-8/S0002-9947-2011-05209-8.pdf
Title: HAUSDORFF MEASURES AND FUNCTIONS OF BOUNDED QUADRATIC VARIATION
Authors: D. APATSIDIS, S. A. ARGYROS, AND V. KANELLOPOULOS
Abstract: To each function $f$ of bounded quadratic variation we associate a Hausdorff measure $\mu_f$ . We show that the map $f \mapsto \mu_f$ is locally Lipschitz and onto the positive cone of $\mathcal{M}[0,1]$. We use the measures $\{\mu_f : f \in V_2\}$ to determine the structure of the subspaces of $V_2^0$ which either contain $c_0$ or the 2 square stopping time space $S^2$.
Chapter 3 is of most interest I think.
Edit
So I think I totally misunderstood Theorem 22 of the paper (I have removed my incorrect interpretation). The theorem states that the set $X:= \{x \in [0,1]: f \text{ is differentiable at } x\}$ has measure 0 under the quadratic variation measure $\mu_f$ (assuming $f$ has finite quadratic variation to begin with). If $f$ is differentiable on $[0,1]$ then we have $X = [0,1]$ and therefore $\mu_f([0,1]) = 0$. This in particular implies the quadratic variation of $f$ is zero (since $\mu_f([0,1])$ equals the quadratic variation). I believe this is Corollary 23 of the paper, which actually proves the stronger claim that if a function of finite quadratic variation is continuous and has only countably many points of non-differentiability then the function has quadratic variation zero. So it looks like the answer is indeed affirmative.