By the Levy zero-one law, the question is equivalent to deciding whether $A$ and $\theta$ are measurable with respect to $\mathcal F_\infty$, the common refinement of all the $\mathcal F_t$. The answer is positive.
For $\alpha>0,\beta\ge 0$, consider
$$Z_n=Z_n(\alpha,\beta):=Y_{\beta+(2n+1)\alpha }-Y_{\beta+(2n-1)\alpha }=U_n+X_n$$
where, for $n=0,1,2,\ldots$, the sequence
$$U_n:=A\sin\Bigl(\theta (\beta+(2n+1)\alpha)\Bigr)-A\sin\Bigl(\theta (\beta+(2n-1)\alpha)\Bigr)=2A\sin(\alpha \theta) \cdot \cos(\theta\beta+2n\theta\alpha)$$
is an almost periodic sequence obtained as a function of the (zero entropy) rotation by angle $2\alpha\theta$, and
$$X_n:=W_{\beta+(2n+1)\alpha }-W_{\beta+(2n-1)\alpha }$$
is a sequence of i.i.d. Gaussian variables with mean 0 and variance $2\alpha$.
In [1], H. Furstenberg defined disjointness of dynamical systems and proved in Theorem I.2 that i.i.d. processes and zero-entropy processes are disjoint.
In Theorem 1.5 of the same paper, he showed that if $(U_n)_{n \ge 1}$ and $(X_n)_{n \ge 1}$ are sequences
of integrable real random variables which define two disjoint stationary
processes, then the sum sequence $(U_n+X_n)_{n \ge 1}$ determines $(U_n)_{n \ge 1}$.
A generalization is in Theorem 7 of [2].
In particular, from the sequence $\{Z_n(\alpha,\beta)\}_{n \ge 1}$ we can obtain
$(U_n)_{n \ge 1}$ =$(U_n(\alpha,\beta))_{n \ge 1}$ for every $\alpha$ and a.e. $\beta$, and using continuity, for all $\beta$. Then $\theta=\pi/(4\alpha_*)$, where
$\alpha_*=\min\{\alpha>0: U_1(\alpha,0)=0\}$ and then $A$ is easy to determine.
Theorem 1.6 in [3] gives another proof.
For an elementary direct argument,
define the deterministic function $f_0(\alpha)=\lim_n \frac{1}{n}\sum_{k=1}^n Z_n(\alpha,0)^2$
and observe that the smallest positive $\alpha$ where $f_0(\alpha)\alpha^{-2}$ is minimized
is $\alpha_{min}=\pi/\theta$. Moreover, $$f_0(\alpha_{min}/2)=4A^2+\alpha_{min}$$
which yields $A$.
[1] H. Furstenberg (1967), Disjointness in ergodic theory, minimal sets, and a problem in
Diophantine approximation, Math. systems theory 1, pp. 1-49.
https://mathweb.ucsd.edu/~asalehig/F_Disjointness.pdf
[2] Furstenberg, Hillel, Yuval Peres, and Benjamin Weiss. "Perfect filtering and double disjointness." In Annales de l'IHP Probabilités et statistiques, vol. 31, no. 3, pp. 453-465. 1995.http://www.numdam.org/article/AIHPB_1995__31_3_453_0.pdf
[3] Lev, Nir, Ron Peled, and Yuval Peres. "Separating signal from noise." Proceedings of the London Mathematical Society 110, no. 4 (2015): 883-931.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.749.2820&rep=rep1&type=pdf
Best Answer
If $F_t(\omega)=\int_0^t h_s(\omega)\,ds$, with $h$ progressive and such that $\int_0^1 h_s^2(\omega)\,ds\le C$ for some finite constant $C$, then indeed $T^*_F\Bbb P$ is equivalent to $\Bbb P$.
Let $X_t$ denote the coordinate process on $\Omega$, a Brownian motion under $\Bbb P$. Now define the martingale $M_t:=\int_0^t h_s\,dX_s$ and the associated exponential martingale $L_t:=\exp(-M_t-{1\over 2}\langle M\rangle_t)$. (This is a martingale by Novikov's criterion.) Finally define $\Bbb Q$ to be the probability measure on $\Omega$ with density $L_1$ with respect to $\Bbb P$. By Girsanov's theorem, the process $Y_t:=X_t+F_t$ is a $\Bbb Q$-Brownian motion. Then for any bounded measurable $f:\Omega\to\Bbb R$, $$ \int_\Omega f\,d\Bbb P = \int_\Omega f(Y)\,d \Bbb Q= \int_\Omega L_1 f(Y)\,d \Bbb P. $$ Thus, $\int_\Omega f\,d\Bbb P=0$ iff $\int_\Omega L_1 f(Y)\,d \Bbb P=0$, and in turn iff $\int_\Omega f\,dT^*_F\Bbb P=\int_\Omega f(Y)\,d \Bbb P=0$ because the density $L_1$ is strictly positive and finite a.s. $\Bbb P$.
The boundedness condition on $h$ can be partially relaxed by localization. Suppose $\int_0^1 h^2_s\,ds<\infty$, $\Bbb P$-a.s. Define, for each positve integer $n$, $T_n:=\inf\{t: \int_0^t h_s^2\,ds>n\}$, and then $F^{(n)}_t:=\int_0^{t\wedge T_n} h_s\,ds$. By the preceding paragraph, $\Bbb P$ is equivalent to $T^*_{F^{(n)}}\Bbb P$ for each $n$. But $F^{(n)}=F$ on the event $G_n:=\{\int_0^1 h^2\,ds\le n\}$. Because $\Bbb P(\cup_nG_n)=1$, it follows that $T^*_F\Bbb P \ll\Bbb P$.