The result holds for any bounded function $f$, in the following sense: for any real $s>1/2$,
\begin{equation}
P^*(A)=0,
\end{equation}
where
\begin{equation}
A:=\Big\{\exists t_0\in[0,1]\ \limsup_{t\to t_0}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}<\infty\Big\},
\end{equation}
$P^*$ is the outer probability, $\limsup_{t\to t_0}:=\limsup_{t\to t_0,t\in[0,1]}$,
\begin{equation}
W_f:=W+f,
\end{equation}
and $W$ is a standard Wiener process.

The proof is obtained by a straightforward adaptation of the proof of the Paley--Wiener--Zygmund theorem on the almost sure nowhere differentiability of the Brownian motion.

Indeed, since $W$ and $f$ are bounded,
\begin{equation}
A\subseteq B=\bigcup_{M=1}^\infty B_M,
\end{equation}
where
\begin{equation}
B:=\Big\{\exists t_0\in[0,1]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}<\infty\Big\},
\end{equation}
\begin{equation}
B_M:=\Big\{\exists t_0\in[0,1]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}\le M\Big\}.
\end{equation}
Next,
\begin{equation}
B_M:=B_{M,1}\cup B_{M,2},
\end{equation}
where
\begin{equation}
B_{M,1}:=\Big\{\exists t_0\in[0,1/2]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}\le M\Big\},
\end{equation}
\begin{equation}
B_{M,2}:=\Big\{\exists t_0\in[1/2,1]\ \sup_{t\in[0,1]}\frac{|W_f(t)-W_f(t_0)|}{|t-t_0|^s}\le M\Big\}.
\end{equation}

Let $r$ be any integer such that
\begin{equation}
r>1/(s-1/2).
\end{equation}
Let then $n$ be any integer $\ge n_r$, where $n_r$ is the smaller integer $q$ such that $2^{q-1}>r$.

Assuming the event $B_{M,1}$ occurs, let $K$ be a random integer in the set $\{1,\dots,2^{n-1}\}$ such that $t_0\in[\frac{K-1}{2^n},\frac K{2^n}]$. Then for $j=1,\dots,r$
\begin{equation}
\begin{aligned}
&\Big|W_f\Big(\frac{K+j}{2^n}\Big)-W_f\Big(\frac{K+j-1}{2^n}\Big)\Big| \\
&\le\Big|W_f\Big(\frac{K+j}{2^n}\Big)-W_f(t_0)\Big|
+\Big|W_f\Big(\frac{K+j-1}{2^n}\Big)-W_f(t_0)\Big| \\
&\le M\Big|\frac{K+j}{2^n}-t_0\Big|^s
+M\Big|\frac{K+j-1}{2^n}-t_0\Big|^s\le\frac{2Mj^s}{2^{sn}} \le\frac{2Mr^s}{2^{sn}}
\end{aligned}
\end{equation}
So,
\begin{equation}
B_{M,1}\subseteq\bigcap_{n\ge n_r}\bigcup_{k=1}^{2^{n-1}}C_{n,k},
\end{equation}
where
\begin{equation}
C_{n,k}:=\bigcap_{j=1}^r \Big\{\Big|W_f\Big(\frac{k+j}{2^n}\Big)-W_f\Big(\frac{k+j-1}{2^n}\Big)\Big|
\le\frac{2Mr^s}{2^{sn}}\Big\}.
\end{equation}
By the independence of the increments of the Wiener process and because the pdf of the normally distributed random variable $W_f\big(\frac{k+j}{2^n}\big)-W_f\big(\frac{k+j-1}{2^n}\big)$ is $\le2^{n/2-1}$, we have
\begin{equation}
\begin{aligned}
P(C_{n,k})&=\prod_{j=1}^r P\Big(\Big|W_f\Big(\frac{k+j}{2^n}\Big)-W_f\Big(\frac{k+j-1}{2^n}\Big)\Big|
\le\frac{2Mr^s}{2^{sn}}\Big) \\
& \le\Big(2^{n/2}\frac{2Mr^s}{2^{sn}}\Big)^r=\frac C{2^{(1+a)n}},
\end{aligned}
\end{equation}
where $C:=(2Mr^s)^r$ and $a:=r(s-1/2)-1>0$. So,
\begin{equation}
P\Big(\bigcup_{k=1}^{2^{n-1}}C_{n,k}\Big)\le2^{n-1}\frac C{2^{(1+a)n}}\le\frac C{2^{an}}
\end{equation}
and
\begin{equation}
P^*(B_{M,1})\le\lim_{n\to\infty}\frac C{2^{an}}=0.
\end{equation}
So, $P^*(B_{M,1})=0$. Similarly, $P^*(B_{M,2})=0$. So, $P^*(B_M)=0$, for all $M$.

Thus,
\begin{equation}
P^*(A)\le P^*(B)\le\sum_{M=1}^\infty P^*(B_M)=0,
\end{equation}
as claimed. $\quad\Box$

Brownian Motionby Peter Mörters and Yuval Peres. I have edited your post accordingly. $\endgroup$