There are general statements on the existence of moments of solutions to SDEs. If you are lucky, you have one of those results at your disposal to deduce that $\sup_{s \leq t} \mathbb{E}(|X_s|^3)<\infty$ for all $t>0$. If this is not the case, then you need to do some further calculations.
In the following, I will consider $\mathbb{E}(X_t^4)$ (rather than $\mathbb{E}(|X_t|^3)$) since this saves us the trouble to deal with the modulus. By Itô's formula, we have
$$X_t^4 = 8 \int_0^t X_s^3 \sqrt{X_s} \, dB_s +30 \int_0^t X_s^3 \, ds.$$
Now we want to take the expectation, but since we do not yet know whether the expectation is finite, we first need to some stopping. To this end, define
$$\tau_r := \inf\{t \geq 0; |X_t| \geq r\},$$
note that $\tau_r \uparrow \infty$ as $r \to \infty$ and $|X_{t \wedge \tau_r}| \leq r$. Since
$$t \mapsto \int_0^{t \wedge \tau_r} X_s \sqrt{X_s} \, dB_s$$
is a martingale (because of the stopping!), and, hence, has expectatio zero, we get
$$\mathbb{E}(X_{t \wedge \tau_r}^4) = 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} X_s^3 \, ds \right).$$
Using the elementary estimate
$$x^3 \leq |x|^3 \leq 1+x^4, \qquad x \in \mathbb{R},$$
for $x=X_s$, we find that
$$\mathbb{E}(X_{t \wedge \tau_r}^4) \leq 30 \mathbb{E} \left( \int_0^{t \wedge \tau_r} (1+X_s^4) \, ds \right) \leq 30 \int_0^t \mathbb{E}(1+X_{s \wedge \tau_r}^4) \, ds.$$
If we define $u(t) := \mathbb{E}(1+X_{t \wedge \tau_r}^4)$ for fixed $r>0$, then
$$u(t) \leq 1+ 30 \int_0^t u(s) \, ds$$
and so Gronwall's lemma yields
$$u(t) \leq c e^{Mt}$$
for suitable constants $c,M>0$, which do not depend on $r>0$. Consequently, we have shown that
$$\mathbb{E}(1+X_{t \wedge \tau_r}^4) \leq c e^{Mt}, \qquad t \geq 0.$$
By Fatou's lemma, this implies
$$\mathbb{E}(1+X_{t}^4) \leq c e^{Mt}, \qquad t \geq 0.$$
In particular,
$$\sup_{t \leq T} \mathbb{E}(X_t^4) < \infty, \qquad T>0,$$
which also yields
$$\sup_{t \leq T} \mathbb{E}(X_t^3) < \infty, \qquad T>0.$$
Suppose $X_t=X_0A_tM_t=X_0A'_tM'_t$, where $A,M,A',M'$ satisfy the given conditions. Then
\begin{align}
\log X_t &= \log X_0 + \log A_t + \log M_t\\
&= \log X_0 + \log A_t + \int_0^t \frac1{M_s}\,dM_s
- \frac12\int_0^t \frac1{M_s^2}\,d\langle M \rangle_s.
\end{align}
Similarly,
$$
\log X_t = \log X_0 + \log A'_t + \int_0^t \frac1{M'_s}\,dM'_s
- \frac12\int_0^t \frac1{(M'_s)^2}\,d\langle M' \rangle_s.
$$
Thus, the local martingale parts of these decompositions agree:
$$
\int_0^t \frac1{M_s}\,dM_s = \int_0^t \frac1{M'_s}\,dM'_s.
$$
The quadratic variation of the above local martingale is therefore
$$
\int_0^t \frac1{M_s^2}\,d\langle M \rangle_s
= \int_0^t \frac1{(M'_s)^2}\,d\langle M' \rangle_s.
$$
Together these show that $\log M_t = \log M'_t$, which implies $M = M'$, and hence, $A = A'$.
Best Answer
Since $[X]_t$ is of finite variation, $\int f_{xx}(s, X_s)d[X]_s$ is interpreted in the Lebesgue-Stieltjes sense. This integral itself produces finite variation functions. You can see that by decomposing $f_{xx}(s, X_s)$ into positive and negative parts and $d[X]_s$ as the difference of two nondecreasing functions.