For fixed $\epsilon>0$ set
$$L_t^{(\epsilon)} := \frac{1}{2} \int_0^t \frac{\epsilon}{(\epsilon+B_s^2)^{3/2}} \, ds,$$
i.e.
$$f_{\epsilon}(B_t) = \sqrt{\epsilon} + \int_0^t \frac{B_s}{\sqrt{\epsilon+B_s^2}} \, dB_s+ L_{t}^{\epsilon}.$$
Since $f_{\epsilon}(B_t) \to |B_t|$ in $L^2$ and the stochastic integral converges in $L^2$ to $\int_0^t \text{sgn}(B_s) \, dB_s$, it follows that
$$L_t^{(\epsilon)} = f_{\epsilon}(B_t)-\sqrt{\epsilon} - \int_0^t \frac{B_s}{\sqrt{\epsilon+B_s^2}} \, dB_s \xrightarrow[L^2]{\epsilon \to 0} |B_t| - \int_0^t \text{sgn}(B_s) \, dB_s =: L_t $$
for fixed $t \geq 0$. Note that $(L_t)_{t \geq 0}$ has (a modification with) continuous sample paths since the stochastic integral has continuous sample paths a.s. (the continuity of $|B_t|$ is obvious from the continuity of $B_t$). Now
$$\sup_{t \leq T} |L_t^{(\epsilon)}-L_t| \leq \sqrt{\epsilon}+ \sup_{t \leq T} |f_{\epsilon}(B_t)-|B_t|| + \sup_{t \leq T} \left| \int_0^t \frac{B_s}{\sqrt{\epsilon+B_s^2}} \, dB_s - \int_0^t \text{sgn}(B_s) \, dB_s \right|.$$
The right-hand side converges to $0$ in $L^2$ as $\epsilon \to 0$, and so does the right-hand side. Convergence in $L^2$ implies that there exists a subsequence converging almost surely, i.e. we can find $\epsilon_k \downarrow 0$ such that $$\sup_{t \leq T} |L_t^{(\epsilon_k)}-L_t| \to 0 \quad \text{a.s.}$$ Since $t \mapsto L_t^{(\epsilon)}$ is increasing, it follows that the limit is also increasing in $t$:
$$L_s = \lim_{k \to \infty} L_s^{(\epsilon_k)} \leq \lim_{k \to \infty} L_t^{(\epsilon_k)} = L_t$$
for any $s \leq t \leq T$.
Remark: Note that we have shown, as a by-product, that
$$|B_t| = \int_0^t \text{sgn}(B_s) \, dB_s + L_t$$
is the Doob-Meyer decomposition of the submartingale $(|B_t|)_{t \geq 0}$.
It may depend on what you mean by "the same". By the occupation time formula for the Brownian local time
$$
\int_0^t{\epsilon\over (\epsilon+B_s^2)^{3/2}}\,ds = \int_{-\infty}^\infty {\epsilon\over(\epsilon+x^2)^{3/2}}L_t^x\,dx=\int_{-\infty}^\infty {1\over (1+u^2)^{3/2}}L^{\sqrt{\epsilon}u}_t\,du.
$$
Because
$$
\int_{-\infty}^\infty {1\over (1+u^2)^{3/2}}\,du=2,
$$
sending $\epsilon$ to $0$ on the right side of (3) therefore yields $L^0_t$ because of the continuity in $x$ of $L^x_t$.
Best Answer
For each pair of rationals $p<q$, your calculation shows that the event $$ A_{p,q}:=\{B \hbox{ is zero free in }[p,q], L_p<L_q\} $$ has probability zero. Thus so does the countable union $A:=\cup_{p<q\in\Bbb Q}A_{p,q}$. As both $L$ and $B$ are continuous on an event $C$ of probability 1, the desired claim holds on $C\setminus A$, and this event has probability 1.