I think this question is a good one, but don't expect an encyclopedic answer — MO is not an encyclopedia. Here are some answers, with the disclaimer that I'm a category theorist but not an algebraic geometer.
To question A, by and large the 21st perspective will probably say that it is definitely a stack, in some notion of the word. Certainly there are applications where you do want to consider the "space" quotient $X/G$, in which points in the same orbit are honestly identified. This is like taking a form of "$\pi_0$" of the stack. (Not etale $\pi_0$, certainly, but a form of $\pi_0$ that's valued in spaces rather than sets.)
To questions B, D, and E, the answer is that, as you guessed, the best definition of $\operatorname{QCoh}(X/G)$ is the category of $G$-modules in $\operatorname{QCoh}(X)$, at least when $G$ is a finite group. The geometric intuition is that a quasicoherent sheaf on $X$ is something like a vector bundle over $X$. In the quotient $X/G$, we add an isomorphism between any two points for each way that they are related by an element of $G$. So a vector bundle over $X/G$ should have a fiber over each point of $X$, and an isomorphism between these fibers for each pair of $G$-related points.
There is a quotient morphism $X \to X/G$. The $\operatorname{QCoh}$ functor is best understood as contravariant, just like $\mathcal{O}$ is contravariant. Namely, a geometric morphisms $f: X \to Y$ correspond (modulo details) to symmetric monoidal "linear" functors $f^\ast : \operatorname{QCoh}(Y) \to \operatorname{QCoh}(X)$, which pull back a "vector bundle" along the map. This is certainly true for $\operatorname{QCoh}(X/G) = \operatorname{QCoh}(X)^G$, with the quotient morphism corresponding to the functor "forget the $G$-action". That said, each such functor $f^\ast$ also has a right adjoint $f_\ast : \operatorname{QCoh}(X) \to \operatorname{QCoh}(Y)$, which is not usually symmetric monoidal — it is that takes a "vector bundle" over $X$ and makes it into the "vector bundle" over $Y$ whose stock over $y\in Y$ is the space of sections over $f^{-1}(y)$ of the corresponding bundle on $X$. In the case of the quotient map $X \to X/G$, its right adjoint is the "free" functor, assigning to a quasicoherent module $M$ the corresponding free $G$-module $G \otimes M$. You ask for $f_\ast$ to be "a morphism of abelian categories", which is vague to me. The best definition I know of "morphism of abelian categories" is a right-exact functor (if I have left and right correct), in which case in general pushforward maps are not morphisms — they are instead left-exact. I think that if $G$ is finite, then in fact the pushforward along the quotient map is exact; maybe I need to include that the characteristic of the ground field does not divide the order of $G$.
As for C, as just a subcategory, I'm sure the answer is yes. If you ask for more conditions, the answer is probably still yes, at least in the finite-group case: you should be able to take the category of cocommutative coalgebras in $\operatorname{QCoh}(X/G)$ and find this as sheaves-of-sets on something. But I'd have to think more about details.
They do not exist in general. The simplest example is maybe to take $X\rightarrow Y$ to be the closed embedding of the origin inside $\mathbb{A}^1.$ Then $f_*$ sends a vector space $V$ to the $\mathcal{D}_{\mathbb{A}^1}$-module $V\otimes\delta_0$, where $\delta_0$ is the irreducible $\mathcal{D}_{\mathbb{A}^1}$-module set-theoretically supported at $0$. The point is that because $\delta_0$ is infinite-dimensional as a vector space, this functor does not commute with products and hence cannot have a left adjoint.
Best Answer
No, not always.
In
Positselski, Leonid; Schnürer, Olaf M., Unbounded derived categories of small and big modules: is the natural functor fully faithful?, J. Pure Appl. Algebra 225, No. 11, Article ID 106722, 23 p. (2021). ZBL1464.18015.
the authors show in Theorem 3.1 that the natural functor $D(\operatorname{mod}R)\to D(\operatorname{Mod}R)$ is not fully faithful when $R$ is a quasi-Frobenius ring of infinite global dimension (such as $R=\mathbb{Z}/4\mathbb{Z}$).
They also have some positive results. For example, Corollary 5.12 states that $D(\operatorname{coh}X)\to D(\operatorname{Coh}X)$ is fully faithful if $X$ is a regular Noetherian scheme of finite Krull dimension.