Does there exist a function $f:\mathbb R\to\mathbb R$ such that $$ \lim_{s\downarrow 0} \frac{f(t+s)-f(t)}{s^\beta}=c,\quad \text{for every }t\in\mathbb R, $$ where $\beta\in(0,1)$ and $c\in\mathbb R\backslash \{0\}$?
I am happy if you answer the same question but with a non-zero continuous function $c(t)$ on the right hand side.
Assume $c>0.$ This condition would imply that the upper and lower right Dini derivatives
$$D^+f(t) =\limsup_{s \to {0+}} \frac{f(t + s) - f(s)}{s}$$ $$D_+f(t) =\liminf_{s \to {0+}} \frac{f(t + s) - f(s)}{s}$$ are $+\infty$ everywhere in $[0,1].$ This contradicts the Denjoy-Young-Saks theorem, which says that the set of $t\in[0,1]$ satisfying $D^+f(t)=D_+(t)=+\infty$ has measure zero.
For a cheaper proof, first note that $f$ is right-continuous and hence measurable. Let $\lambda=2^{(1-\beta)/2}>1$ and define sets
$$S_\epsilon=\{t\in[0,1]\mid f(t+\epsilon)>f(t)+c\epsilon^\beta/\lambda\text{ and }f(t+2\epsilon)<f(t)+c(2\epsilon)^\beta \lambda\}.$$
Each $t\in[0,1]$ will be in $S_\epsilon$ for all sufficiently small $\epsilon,$ so for some $\epsilon<\tfrac13$ we have $\mu(S_\epsilon)>\tfrac 2 3.$ The translate $S_\epsilon-\epsilon$ then has measure at least $\tfrac13$ in $[0,1],$ so there is a point $t\in S_\epsilon\cap(S_\epsilon-\epsilon).$ This gives
$$f(t)+2c\epsilon^\beta/\lambda<f(t+\epsilon)+c\epsilon^\beta /\lambda<f(t+2\epsilon)<f(t)+c(2\epsilon)^\beta \lambda$$ which contradicts $\lambda=2^{(1-\beta)/2}.$
The set of points that verify the condition is in fact countable, whence of Hausdorff dimension zero (and in particular of measure zero as already shown by Dap).
Without loss of generality assume $c>0$. As $\beta<1$ we may choose $c_1<c<c_2$ so that $c_2< 2^{1-\beta}c_1$. We define for $t\in {\Bbb R}$: $$ \Delta_t = \sup \{ \delta>0: c_1s^\beta \leq f(t+s)-f(t)\leq c_2 s^\beta, \;\;\forall \; 0 \leq s\leq \delta \} $$ and then for $n\geq 1$: $$ \Omega_n = \{ t\in {\Bbb R} : \Delta_t>\frac{1}{2^n} \}$$ A point $t\in {\Bbb R}$ verifying the condition must belong to some $\Omega_n$. We will show that each $\Omega_n$ is countable, proving our claim.
So pick $t\in \Omega_n$ and $0<\epsilon < \frac{1}{2^{n+1}}$. Assume that $t_1=t+\epsilon\in \Omega_n$. Then with $s=\epsilon$ we should have: $$ c_1 s^\beta \leq f(t_1+s)-f(t_1) =(f(t_1+s)-f(t))-(f(t_1)-f(t)) \leq c_2(\epsilon+s)^\beta-c_1\epsilon^\beta$$ or with $s=\epsilon$ the inequality: $\;2c_1 \epsilon^\beta \leq c_2 (2\epsilon)^\beta < 2c_1 \epsilon^\beta$, a contradiction. Thus, points in $\Omega_n$ have to be at least $2^{-(n+1)}$ separated and the conclusion follows.
If $c$ is continuous or even discontinuous (but finite and non-zero) the same conclusion follows, since you may cover the possible $c$ values by countable many intervals of the above type $(c_1,c_2)$ with $c_2<c_12^{1-\beta}$ and then for each such intervals only countable many points may verify the corresponding criterion.
