Here's how you can proceed once you've applied the Kurosh subgroup theorem.
Starting from the expression $H = (*_i H_i) * F$ for your subgroup $H \le G$, break the first term into
$$(*_{i \in I_0} H_i) * (*_{i \in I_1} H_i) * (*_{i \in I_2} H_i) * F
$$
where if $i \in I_k$ then $H_i$ is a free abelian group of rank $k$ ($k = 0,1,2$).
We can then collect terms of the free product and rewrite this as
$$(*_{i \in I_2} H_i) * \underbrace{\left(F * (*_{i \in I_1} F_i) \right)}_{\text{free of rank $R_1 = \text{rank}(F) + \left| I_1 \right|$}}
$$
So what you have is a free product of $R_2 = \left| I_2 \right|$ rank 2 free abelian groups and a free group of rank $R_1$. Using Grushko's Theorem, you can prove that such a group is determined, up to isomorphism, by the the ordered pair $(R_1,R_2)$.
So, your subgroup $H$ is isomorphism to the whole group $G$ if and only if $\left|I_2\right|=2$ and $\text{rank}(F) = \left|I_1\right|=0$.
Since your subgroup $H$ has finite index in $G$, the quotient graph of groups $T/H$ is a finite tree. Since a vertex of $T/H$ labelled with the trivial group would have to have countably infinite valence, there can be no such vertices. And there are no vertices labelled with a rank $1$ abelian group. All remaining vertices of $T/H$ are therefore labelled with rank $2$ abelian groups. Since there are only two such vertices, the the graph of groups $T/H$ is forced to be of exactly the same type as $T/G$: two $\mathbb Z^2$ vertices and one edge labelled by the trivial group. It follows that any single edge $E$ of the original Bass-Serre tree is simultaneously a fundamental domain for the whole group $G$ and for its subgroup $H$. Since $G$ acts freely on edges, it follows that $H=G$.
The answer to 1 is: almost yes. The "almost" is needed because of the possibility that there is an edge $e$ and a finite order element $g \in G$ such that $ge = \bar e$, in which case $g$ does not fix any vertex. The solution to that is to subdivide each edge $e$ at its midpoint, if there exists such a $g$. Having done that, one reduces to the case under the action of $G$ on $T$, no such edge $e$ exists, and now the answer is an unqualified yes. The proof is that every finite subgroup must fix a point, and therefore it must fix a vertex.
The answer to 2 is no, and it's still no for example where all edge groups are trivial. I think that you mis-applied Grushko's theorem, as I'll explain below.
For a simple counterexample, take $\Gamma$ to be the rose graph with $n$ petals, and take each $G_v$ and $G_e$ to be the trivial group. In this case $G$ is a rank $n$ free group, and the petals represent a particular free basis of $G$. But $\text{Out}(G)$ is a large and complicated group in that situation, and there are many elements which are not represented by any graph isomorphism as you suggest.
Similar counterexamples with nontrivial vertex groups are easily obtained. For example, just take the same rose graph $\Gamma$ but take $G_v$ to be your favorite finite group. In this case $\text{Out}(G)$ is still a large and complicated group that does not preserve the splitting represented by $\Gamma$.
As for Grushko's Theorem applied in the case that the edge groups are trivial, that theorem only tells you that an automorphism permutes the vertex groups, it makes no guarantee about preserving the splitting itself represented by $\Gamma$.
Added after comments: Although it remains unclear to me exactly what is the intent of your question 2, let me add a few remarks.
One thing I can think of to say is that after properly formulating an equivalence relation on the set of actions of $G$ on simplicial trees, there is a natural action of the group $\text{Out}(G)$ on the set of equivalence classes. One could equivalently describe this action with a proper formulation of an equivalence relation on the set of graph-of-groups descriptions of $G$. This action can be roughly described as in your alternative answer.
This was first done for the case of a finite rank free group $G=F_n$ and free actions on trees, in the Culler-Vogtmann paper "Moduli of graphs". Much more was done in that paper, including applications to the group $\text{Out}(F_n)$, by packaging these equivalence classes into a kind of "deformation space" for a free group (as we now think of it more generally; see below), known as the outer space of the free group.
The completely general theory was done in the Guirardel-Levitt paper "Deformation spaces of trees".
Best Answer
As I mentioned in the comments, free products give you actions on trees with the properties you desire.
I wrote out a description of the Bass-Serre tree for a free product $A*B$ and the associated action here.
More exotic examples can be cooked up using HNN-extensions or free products with amalgamation (e.g. finite but non-trivial edge stabilisers in (1)).