Firstly, at the beginning of the question you are missing irreducibility/cuspidality assumptions. If $\rho_1$ and $\rho_2$ are $\ell$-adic Galois reps with the same char poly in a set of primes of density 1, then you can only deduce their semisimplifications are isomorphic. A counterexample to your statement would be given by $\rho_1=1+\omega$ ($\omega$ the cyclotomic character) and $\rho_2$ some non-split extension coming from Kummer theory (taking $\ell$-poower roots of some prime number $p$ for example). Of course you knew that already.

On the automorphic side you make "the same slip", well, a related slip. If $\chi$ and $\psi$ are two Grossencharacters and their ratio is the norm character at some place $v$ (or possibly even infinitely many, or even all $v$), then the induction of $\chi+\psi$ from the Borel to $GL_2(\mathbf{A})$ is reducible, and any Jordan-Hoelder factor (which by definition means a tensor product of J-H factors of the local inductions) ramified at all but finitely many places is an automorphic representation. So now you can build two non-cuspidal automorphic representations which are isomorphic at all but one place (and such that the local components don't even have the same dimension at the bad place) easily. Of course for cuspidal representations you have strong multiplicity 1 theorems.

Passing remark: it is a source of some confusion to me as to why these errors are "similar" but on the other hand they don't "biject". The problem is that on the Kummer theory side, if you allow ramification at two primes, you get a whole projective line of non-isomorphic Galois representations (all with the same semisimplificiation). On the automorphic side the amount of control you have is much more combinatorial (you can change a finite set of places from trivial to Steinberg and that's it). So $\pi$ and $\rho$s don't match up (so Toby Gee and I only conjecture that given any (EDIT: algebraic) $\pi$ one expects a semi-simple $\rho$, and nothing more, and for $GL_n$ this observation is no doubt much older).

OK so after these pedantic remarks, that you no doubt knew anyway, but could have avoided if you'd put "irreducible" and "cuspidal" in the appropriate places, we move on to the far more interesting question of whether one can do better than multiplicity 1. And your hunch is correct. First you should do the following exercise on the Galois side: if $\rho_1$ and $\rho_2$ are two irreducible 2-dimensional representations of a finite group $G$, and their traces agree on a set $S$ in $G$ of density greater than 7/8, then $\rho_1$ and $\rho_2$ are isomorphic. Big hint: orthogonality relations. Next you should convince yourself that 7/8 is optimal in this result Big hint: D_8 x D_8. (EDIT: slightly simpler is D_8 x C_2---here D_8 has 8 elements). So now for silly Artin representation reasons you can't expect to beat 7/8 (like you can't expect to beat 1/2 in the GL_1 case as in Jared's comment).

So now the great news is that Dinakar Ramakrishnan proved an analogue on the automorphic side! At least in some cases. See the appendix to "$l$-adic representations associated to modular forms over imaginary quadratic fields. II." by Richard Taylor (Inventiones 116).

As for your second question though, it's completely hopeless. Even for $GL_1$ your hope is (provably) way out, so for $GL_2$ one can perhaps construct "Eisenstein" counterexamples (induced from non-automorphic characters of $GL_1$). The first objection is that $\pi_p$ can be ramified for infinitely many $p$, making you dead in the water. But even if $\pi_p$ is unramified for all $p$ you've still got no chance. Let me stick to $GL_1/\mathbf{Q}$. The point is that every Grossencharacter for $GL_1/\mathbf{Q}$ is the product of a finite order character $\chi$ (a Dirichlet character) and $||.||^s$, so you can recursively construct characters of $\mathbf{Q}_p^\times$ each of which is totally at odds with everything that came before. For example lets send $Frob_2$ to 1. Now let's write down all the solutions to $2^s.\zeta=1$, with $\zeta$ a root of unity and $s$ a complex number. There are only countably many values of $s$ in this list. So choose $s$ not in the list and send $Frob_3$ to $3^s$. Now knock off countably many more $s$ and send $Frob_5$ to $5^s$ for $s$ in neither list. We are making a collection of representations here that are pairwise completely incompatible! This is only one of the many obstructions. For example, for cuspidal automorphic representations there are Weil bound obstructions (Ramanujan conjecture) and arithmeticity obstructions in the holomorphic case---all deep theorems or conjectures about automorphic forms in some cases that can easily be violated if we're allowed to build $\pi$ locally. These arguments trivially show that the set of automorphic $\pi$s have density zero (and "a very small zero" at that) amongst all the $\pi$s. In fact here's a much cleaner objection for $GL_2$: if you stick to cuspidal automorphic representations with a fixed central character then there are only countably many! But you have uncountably many choices at each local place! So it's completely hopeless I think.