The issue here is precisely what you mean when you say "regular path $J_t$ of almost complex structures".
I am not particularly expert in the theory of holomorphic curves, so I won't comment on that; I expect the situation is precisely analagous to the finite-dimensional setting I will describe below (I'm sure there are more technical issues involving the corner points).
Let $M$ be a smooth manifold, and $S \subset N$ a closed submanifold. Given a map $f: M \to N$ transverse to $S$, we have that $\mathcal M_f = f^{-1}(S)$ is a smooth manifold (with corners etc if $S$ and $N$ have them).
The condition "$f \pitchfork S$" is the precise analogy to "$J_t$ is regular" in this finite-dimensional setting.
Suppose you have a path of functions $f$, written as a map $f_t: I \times M \to N$. Then I would say that this is a regular path of functions if the map $I \times M \to N$ is transverse to $S$, and the same for $\{0,1\} \times M \to N$. Then the "parameterized moduli space" $\mathcal M_I$ is a smooth manifold with boundary equal to $\mathcal M_0 \sqcup \mathcal M_1$. This space is equipped with a smooth map $\mathcal M_I \to I$ which is a submersion near the boundary.
The point is that each individual $f_t$ is not assumed to be regular; spelling it out, it is entirely possible that for $(t,x)$, we have $f_t(x) \in S$, and $\text{Im}(df_t) \subset N_{f_t(x)} S$ is not the entirety of the normal space, but rather of codimension 1; and so that together with the additional vector $\frac{d}{dt} f_t(x) \in N_{f_t(x)} S$, we span the whole of the normal space.
The second point is that an individual $\mathcal M_t$ is regular if and only if $t \in I$ is a regular value of the projection $\mathcal M_I \to I$.
So if, in addition, $f_t$ is regular for each $t$, then by definition the projection $\mathcal M_I \to I$ is a submersion; as you know well, a proper smooth submersion is a fiber bundle, and this provides a global diffeomorphism $\mathcal M_I \cong I \times \mathcal M_0$.
In almost every interesting application, given any two regular functions / almost complex structures / whatever, it will be essentially impossible to find a path through regular (blahs) between them; but it will be possible to find a "regular path" of (blahs) between them in the sense given above. A simple case to think about the discussion above with is $S$ a hypersurface and $M$ a point! If $S$ disconnects the codomain, then to get from one side to another you will need to cross $S$. In this situation, $\mathcal M_I$ will be a finite set of points which project to the interior of $I$.
Although you are correct in stating that the two differentiable structures are different in so far as they are different sets of coordinate charts, they are in a more abstract sense the same. What I mean by this is that a diffeomorphism allows you to identify whatever differentiable structure one manifold has with that of the other, so the structures are in fact equivalent. This is just like isomorphisms in algebra or homeomorphisms in topology.
The fact that two manifolds are diffeomorphic entails that these two manifolds will share properties pertaining to differentiability, just as two isomorphic vector spaces will share linear-algebraic properties such as dimension or two homeomorphic topological spaces will share topological properties such as compactness. In essence, from the differential-topological point of view, these two manifolds are the same, just with different names for points and charts.
To answer your second question, no, having a diffeomorphism is much stronger than being able to map smoothly from one manifold to another, but yes, the diffeomorphism cannot be the identity.
Regarding your last question, not quite. Rather, this situation resembles the following:
Let $\mathcal{T}_1$ be the topology on $\mathbb{R}$ generated by all sets of the form $(a,b]$ and $\mathcal{T}_2$ be the topology generated by sets of the form $[a,b)$. These topologies have different open sets, but the map $x \mapsto -x$ is a homeomorphism, so every topological property (e.g. countability/separation axioms, compactness, etc) that holds for $(\mathbb{R}, \mathcal{T}_1)$ also holds for $(\mathbb{R}, \mathcal{T}_2)$.
Actually, it is an interesting exercise to show that there is only one smooth structure on $\mathbb{R}$ up to diffeomorphism, so no matter how you define a smooth structure on $\mathbb{R}$ it will be essentially the same as the standard one.
Best Answer
Think of your surface as a balloon in ambient space. Then you can flatten it locally by pressing it on a table. Then this gives you a smooth identification with the plane of the table, at least locally.