Here is an example of a nonconnective $E_\infty$ ring spectrum which, I think, illustrates a key problem. (A more extensive discussion of this phenomenon occurs in Lurie's DAG VIII and in a paper by Bhatt and Halpern-Leinster.)
Let $R$ be an ordinary commutative ring, viewed as an $E_\infty$ ring concentrated in degree zero, and $A$ be the homotopy pullback / derived pullback in the following diagram of $E_\infty$ rings:
Then $\pi_0 A = R[x,y]$ and $\pi_{-1} A$ is the local cohomology group $R[x,y] / (x^\infty, y^\infty)$; all the other homotopy groups of $A$ are zero. As a result, there's a map $R[x,y] \to A$ of $E_\infty$ rings, and any $A$-module becomes an $R[x,y]$-module by restriction.
Here's a theorem. The forgetful map from the derived category $D(A)$ to the derived category $D(R[x,y])$ is fully faithful, and its essential image consists of modules supported away from the origin. This extends to an equivalence of $\infty$-categories.
We could think of this in the following way. The ring $A$ is the $E_\infty$ ring of sections $\Gamma(\Bbb A^2 \setminus \{0\}, \mathcal{O}_{\Bbb A^2})$ on the complement of the origin in affine 2-space over $R$, and the above tells us we actually have an equivalence between $A$-modules and (complexes of) quasicoherent sheaves on $\Bbb A^2 \setminus \{0\}$.
Here are some takeaways from this.
Nonconnective ring spectra are actually quite natural. Global section objects $\Gamma(X, \mathcal{O}_X)$ are usually nonconnective, and we're certainly interested in those.
The above says that even though the punctured plane is not affine but merely quasi-affine, it becomes affine in nonconnective DAG. This is a general phenomenon.
Solely on the level of coefficient rings, the map $R[x,y] \to A$ looks terrible. It is indistinguishable from a square-zero extension $R[x,y] \oplus R[x,y]/(x^\infty,y^\infty)[-1]$. (There is more structure that does distinguish them.)
Many of the definitions as given in DAG for a map are given in terms of the effect (locally) of a map $B \to A$ of ring spectra (e.g. flatness, étaleness, etc etc). For connective objects, this works very well. However, we have just shown that for nonconnective objects, a map of ring spectra may have nice properties—the map $Spec(A) \to Spec(R[x,y])$ should be an open immersion!—which are completely invisible on the level of coefficient rings. This goes for the rings themselves and doubly so for their module categories.
If I have one point here, it is that trying to give definitions in nonconnective DAG in terms of coefficient rings is like trying to define properties of a map of schemes $X \to Y$ in terms of the global section rings $\Gamma(Y,\mathcal{O}_Y) \to \Gamma(X,\mathcal{O}_X)$. This makes nonconnective DAG fundamentally harder.
So far as your question A, this places me somewhere in between your two options (1) and (2). I don't think (1) is right because I think that nonconnective objects are much too important; I have a mild objection to the language in (2) because I don't think that nonconnective objects are straightforward.
So the issue is with this:
all of the groups occuring in the Goerss--Hopkins obstruction theory
vanish
In "generic terms", for the obstruction theory that you're running in either the $K(1)$-local case or the rational case you care about some bigraded obstruction groups
$$
\mathfrak{E}xt^{s,t}(R;S)
$$
where these are some nonabelian Ext-groups occuring in some algebraic category ($K(1)$-locally it is Andre-Quillen cohomology for derived $p$-complete $\psi$-$\theta$-algebras, and rationally it is Andre-Quillen cohomology for graded-commutative algebras over $\Bbb Q$, maybe relative to some fixed base algebra, maybe over some other algebra, maybe on a category of diagrams with some sheaf condition, etc etc etc).
The obstructions to existence of some object typically occur when $t-s = -1$, and to uniqueness (up to homotopy equivalence) occur when $t-s = 0$. These are the groups that vanish when constructing $Tmf$ by obstruction theory. Those two groups, however, tell you only about path components. In general, there is some obstruction-theoretic yoga that tells you $\mathfrak{E}xt^{s,t}(R;S)$ is part of some kind of fringed spectral sequence, computing $\pi_{t-s} \mathcal{M}$ of some moduli space of realizations. As a result, for contractibility of this moduli space we would need to know about what happens to $\mathfrak{E}xt^{s,t}$ for $t-s > 0$, and those groups don't vanish.
(Aside, the Goerss--Hopkins--Miller theorem about Lubin-Tate theories is so strong precisely because, in that situation, all of these higher groups vanish, and so these moduli spaces are homotopy discrete.)
The rational case makes it easiest to see this. For the rational case, roughly (ignoring the sheaf stuff), we have constructed $p$-complete $Tmf_p$ and we need to construct a map $H\Bbb Q[c_4, c_6] \to (Tmf_p)_{\Bbb Q}$ to the rationalization that has a particular effect on $\pi_*$. That map is unique up to homotopy. However, the space of maps $H\Bbb Q[c_4, c_6] \to (Tmf_p)_{\Bbb Q}$ does not have contractible components: $H\Bbb Q[c_4, c_6]$ is a free rational commutative algebra on generators in degrees $8$ and $12$, and so the space of maps to $(Tmf_p)_{\Bbb Q}$ is homotopy equivalent to
$$
Map(S^8, (Tmf_p)_{\Bbb Q}) \times Map(S^{12},(Tmf_p)_{\Bbb Q}).
$$
Its higher homotopy give many contributions from the rational higher homotopy of $Tmf_p$. In this case, freeness actually gives us vanishing of Ext groups when $s > 0$; it's $t$ that gives us trouble.
Best Answer
Yes, this has been achieved in some sense. There is a (unpublished and possibly not yet written) work of Gaitsgory and Lurie where they propose an answer to this question. Given a split reductive group scheme $G$ over $Spec(\mathbb{Z})$ they use a version of the geometric Satake correspondence to construct a stable infinity category which is linear over the sphere spectrum and base changes to the category of representations of $G$. The idea is to look at Whittaker sheaves of spectra over the affine Grassmanian for the complex Langlands dual group. This almost gives you what you want. A small caveat is that the convolution product only gives a braided monoidal structure on this category of spectral representations and they expect that the braiding is unavoidable. So there is no cheap Tannaka duality answer but you get as close to it as you can.