= Solution
Fix a reference path $\gamma_*$ from $L_0$ to $L_1$. By path connectedness, a path $\gamma$ can be joined to it by a smooth homotopy strip $u:[0,1]_s\times[0,1]_t\to M$ with
$$
u(0,t)=\gamma_*(t),\quad u(1,t)=\gamma(t),\quad u(s,0)\in L_0,\quad u(s,1)\in L_1.
$$
Smooth homotopies may be used by the relative smoothing theorem; piecewise smooth homotopies give the same integrals. Orient the square by $ds\wedge dt$ and define the <Lagrangian path-area functional> by
$$
\boxed{\mathcal A(\gamma)=-\int_{[0,1]^2}u^*\omega.}
$$
Here $\mathcal A$ denotes the functional called $A$ in the question. The minus sign is the convention for which downward <gradient flow> is $J$-holomorphic on the standard oriented strip.
For <relative symplectic-area independence>, glue two choices, with one orientation reversed, along their common reference and final paths. The resulting oriented surface is a relative $2$-cycle $Z$ in $(M,L_0\cup L_1)$. Since $H_2(M,L_0\cup L_1)=0$, it bounds a relative $3$-chain: as an absolute chain, $Z=\partial B+C$ for a chain $C$ supported in $L_0\cup L_1$. The <symplectic form> is closed, and it restricts to zero on each <Lagrangian submanifold>. Using piecewise smooth relative chains, <Stokes theorem> gives
$$
\int_Z\omega=\int_Bd\omega+\int_C\omega=0.
$$
Equivalently, integration of the closed <symplectic form> defines the zero pairing on the zero relative <homology> group. The two strips therefore define the same value of $\mathcal A$.
Choosing a different reference path changes every value by the negative area of one fixed connecting strip, independent of $\gamma$. Changing the initially assigned reference value also adds a constant. Path connectedness ensures that this is \b[one global constant], rather than an independent constant on each component.
For the first variation, let $\xi(t)$ be a <variation vector field> along $\gamma$, with $\xi(0)\in T_{\gamma(0)}L_0$ and $\xi(1)\in T_{\gamma(1)}L_1$. Differentiate the defining strip integral. <Stokes theorem>, or the variation formula for the integral of a closed <differential form>, reduces it to the final edge; the side edges give zero because their variations and tangent vectors lie in the <Lagrangian submanifolds>. With the chosen sign,
$$
\boxed{(d\mathcal A)_\gamma(\xi)=-\int_0^1\omega(\xi(t),\dot\gamma(t))\,dt
=\int_0^1\omega(\dot\gamma(t),\xi(t))\,dt.}
$$
If this vanishes for every admissible $\xi$, it in particular vanishes for every field supported in $(0,1)$. Nondegeneracy of the <symplectic form>, together with the <fundamental lemma of the calculus of variations>, forces $\dot\gamma(t)=0$ in the interior and hence everywhere by smoothness. Conversely, a constant path makes the displayed integral zero. The constant must lie in both endpoint submanifolds. Thus
$$
\boxed{\operatorname{Crit}(\mathcal A)=\{\gamma(t)\equiv x:x\in L_0\cap L_1\}.}
$$
No transverse-intersection assumption is required; the critical set can be empty or have positive dimension.
A <compatible almost complex structure> is a smooth bundle map $J:TM\to TM$ such that
$$
J^2=-I,\qquad\omega(Jv,Jw)=\omega(v,w),\qquad\omega(v,Jv)>0\quad(v\ne0).
$$
These conditions make
$$
\boxed{g(v,w)=\omega(v,Jw)}
$$
a <Riemannian metric>. Its symmetry follows from the $J$-invariance and skew symmetry of $\omega$, and its positivity is the final compatibility condition. It also satisfies $g(Jv,w)=\omega(v,w)$.
Use the formal $L^2$ metric on the <path space>,
$$
G_\gamma(\xi,\eta)=\int_0^1g(\xi(t),\eta(t))\,dt.
$$
The variation formula becomes
$$
(d\mathcal A)_\gamma(\xi)=G_\gamma(J\dot\gamma,\xi),\qquad\operatorname{grad}\mathcal A=J\dot\gamma.
$$
Thus a formal downward flow $s\mapsto u(s,\cdot)$ satisfies
$$
\boxed{\partial_su+J(u)\partial_tu=0,\qquad u(s,0)\in L_0,\quad u(s,1)\in L_1.}
$$
Put the standard complex structure $j$ on the strip, with $j\partial_s=\partial_t$. The <J-holomorphic curve> equation $du\circ j=J\circ du$ says $u_t=Ju_s$, which is equivalent to the displayed flow equation because $J^2=-I$. The converse is identical: a smooth <J-holomorphic curve> on the strip with the specified <Lagrangian boundary conditions> gives a formal downward trajectory of $\mathcal A$.
As a sign check, along such a strip,
$$
\frac{d}{ds}\mathcal A(u(s,\cdot))=-\int_0^1|u_s|_g^2\,dt,
\qquad\omega(u_s,u_t)=\omega(u_s,Ju_s)=|u_s|_g^2.
$$
The drop in $\mathcal A$ therefore equals the positive <symplectic area> swept out by the strip. The word formal matters: the $L^2$ calculation gives the interior equation and its <Lagrangian boundary conditions>; it does not assert that arbitrary smooth initial paths produce a well-posed ordinary flow on the smooth <path space>. In particular, the expression $J\dot\gamma$ need not satisfy the endpoint tangent restrictions for an arbitrary initial path. The claimed correspondence concerns smooth solutions of the strip equation.
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