It's an old result by R. Alperin that a compact group has no abstract homomorphism onto $\mathbf{Z}$. [Compact groups acting on trees. Houston J. Math. 6 (1980), no. 4, 439--441.] So Q1 has a negative answer.
Here $\mathbf{Z}$ cannot be replaced by any countable abelian group. For instance, the direct product $G$ of all finite perfect groups (up to isomorphism) has an infinite abstract abelianization. Indeed since $G$ is isomorphic to its countable power, if $G$ were perfect it would be uniformly perfect, and thus all finite groups would be together perfect with a uniform commutator width, which is false. So $G$ has nontrivial abstract abelianization, and again since $G$ is isomorphic to its countable power, its abstract abelianization is uncountable (and thus admits a countable infinite quotient).
About Q2: I don't know any example of a profinite group $G$ and finite simple group $S$ such that $G$ admits $S$ as a quotient abstractly, but not as a quotient by any open subgroup (in all the examples I know with $G$ admitting $S$ as a quotient by a non-open subgroup, $G$ actually has infinitely many open subgroups $H$ with $G/H\simeq S$) .