As suggested in my comment, here's a simple fact which applies to any probability measure $\mu$ on (the Borel σ-algebra of) a second countable topological space $X$. There is a unique minimal closed subset $S$ of $X$ with $\mu(S)=1$ -- the support of $\mu$ -- and, if $X_1,X_2,\ldots$ is an IID sequence of random variables, each with distribution $\mu$, then $\lbrace X_1,X_2,\ldots\rbrace$ is almost surely a dense subset of $S$.
Now, in the situation described in the question, choose a sequence $0\le t_0 < t_1 < t_2 < \cdots \le1$. Then, $X_k\equiv(t_k-t_{k-1})^{-1/2}(B_{t_k}-B_{t_{k-1}})$ is an IID sequence of random variables each with distribution $\mu$. Also, as it is non-degenerate, the support of $\mu$ is not contained in a closed proper subspace of the Banach space $W$.
Claim: Let $0<\alpha<1/2$ and $0<C<\infty$. The random variable $H_{C,\alpha}$ has an absolutely continuous distribution.
Proof: Let $\{{\cal F}_t\}_{t \ge 0}$ be the standard Brownian Filtration. Given a non-negative rational $q$, define the ${\cal F}_q$-measurable random boundary functions $\psi_q^+:[q,\infty) \to (-\infty,\infty)$ and $\psi_q^-:[q,\infty) \to (-\infty,\infty)$ by
$$\psi_q^+(t)=\inf_{s \le q} \; [W_s+C(t-s)^\alpha]$$
and
$$\psi_q^-(t)=\sup_{s \le q} \; [W_s-C(t-s)^\alpha]\,.$$
Observe that both these functions are a.s. continuous, and $\psi_q^+$ is nondecreasing.
For $$q \in G^+=\{q \in {\mathbb Q}: \psi_q^+(q)>W_q\} \,,$$ consider the stopping time
$$H_{C, \alpha}^+(q) := \inf \big \{t \ge q \,: W_t = \psi_q^+(t)\} \,.$$
Similarly, for $$q \in G^-:=\{q \in {\mathbb Q}: \psi_q^-(q)<W_q\} \,,$$ consider the stopping time
$$H_{C, \alpha}^-(q) := \inf \big \{t \ge q \,: W_t = \psi_q^-(t)\} \,.$$
Since $W$ is locally $\beta$-Holder continuous for $\beta \in (\alpha,1/2)$, we infer that with probability 1, we have
$$H_{C,\alpha} \in \{H_{C, \alpha}^+(q) : q \in G^+\} \cup \{H_{C, \alpha}^-(q) : q \in G^-\} \,,$$
so by symmetry, it suffices to prove that for fixed $q \in G^+$, the stopping time
$H_{C, \alpha}^+(q)$ has an absolutely continuous distribution. We deduce this from the next Lemma using the Markov property of $W$ at time $q$. QED
Lemma: Let $\psi:[0,\infty) \to [r,\infty)$ be a nondecreasing function, with $\psi(0)=r>0$. Denote $H:=\inf \{t \ge 0 : W_t=\psi(t)\},$ where the infimum of the empty set is $\infty$.
Then for all $t,\epsilon>0$ and some absolute constant $A$, we have
$$P\Bigl(H \in [t,t+\epsilon]\Bigr) \le Ar^{-2}\epsilon \,.$$
Proof: For $b>0$, let $\tau_b:=\inf \{t \ge 0 : W_t=b\}.$
Then $\tau_1$ has a Levy distribution, with density bounded above by some absolute constant $A>0$. See e.g. https://en.wikipedia.org/wiki/First-hitting-time_model
Since $\tau_b$ has the same law as $b^2 \tau_1$, the density of $\tau_b$ is bounded above by $Ab^{-2}$. Take $b=\psi(t) \ge r$. Then
$$P\Bigl(H \in [t,t+\epsilon]\Bigr) \le P\Bigl(\tau_b \in [t,t+\epsilon]\Bigr) \le Ab^{-2}\epsilon \,,$$
and the lemma follows.
Best Answer
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. $\quad\Box$