The answer is yes, provided that you write your equation in Stratonovich form, rather than Itô form (and assuming that $\mu$ and $\sigma$ are sufficiently smooth in their arguments). The reason is that in one dimension the solution to the Stratonovich equation is a continuous map of $W$ in the sup-norm topology, as observed by Doss in 1977.
This breaks in higher dimensions, but the answer to your question remains the same although I don't know if anyone wrote it up in precisely this way. (Various proofs of the Stroock-Varadhan support theorem use closely related variants of this statement. Note that it is again the Stratonovich formulation which is relevant.)
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
No such $M$ exists for the following $\sigma$. Partition $[0,1]$ into countably many intervals, with endpoints $t_0=0,t_1,t_2,...$ and let $\sigma$ take value 1 on the odd ones and value 2 on the even ones. Given any $M$, the following event $A_M$ has positive probability:
$A_M$ requires that the increments $W_{t_k}-W_{t_{k-1}}$ are in $(9,10)$ for the first $M$ odd values of $k$, and in $(-6,-7)$ for the first $M$ even values of $k$, with $W_t-W_{t_{k-1}}>-1$ for $t \in [t_{k-1},t_k]$ when $0<k<2M$.
Then given $A_M$, we have $W_{t_{2M}}>2M$, so the event $W_1>M$ holds with high probability, yet on $A_M$, the stochastic integral considered is $<-2M$ if you integrate up to $t_{2M}$, and the integral over $[t_{2M},1]$ is likely to be $<2M$.