I think the fundamental object is quantum mechanics is not the Hilbert space and operators on it but the C*-algebra of observables. In this picture the Hilbert space appears as a representation of the algebra. Different irreducible representations are different superselection sectors. The answer to "which operator is observable" is thus simple: the observable operators are those that come from the algebra. Indeed its better to think of observables as self-adjoint elements of the algebra rather than as operators
You might ask where do we get the algebra from. Well, this already should be supplied by the particular model. For a quantum mechanical particle moving on a manifold $M$, the C*-algebra consists of all bounded operators on $L^2(\hat{M})$ commuting with $\pi_1(M)$ where $\hat{M}$ is the universal cover of $M$. The superselection sectors correspond to irreducible representations of $\pi_1(M)$. For QFTs the problem of constructing the algebra of observables is in general open however certain cases (such as free QFT and I believe rational CFT as well) were solved. An approach emphasizing the algebra point of view is Haag-Kastler axiomatic QFT
From the point of view of deformation quantization the quantum observable algebra is a non-commutative deformation of the algebra of continuous (say) functions on the classical phase space. This point of view is not fundamental but it's useful. For example it allows to understand different values of spin and different quantum statistics as superselection sectors
A first remark: the term "Hermitian", even if very popular in physics is in my opinion quite misleading (because someone uses it for symmetric operators, others for self-adjoint ones).
A second remark: the self-adjoint operators of a given Hilbert space $\mathscr{H}$ are in one-to-one correspondence with the strongly continuous groups of unitary operators; not with any group of unitary operators. So it is not possible to associate observables with "unitary operators", but it is possible to associate them with strongly continuous (abelian, locally compact) groups of unitary operators.
These distinctions, even if in some sense subtle, may be important. In fact there are representations of unitary groups that does not admit a self-adjoint generator; for example the canonical commutation relations (in the exponentiated Weyl form) have such "non-regular" representations for fields, and are physically related to infrared problems (see e.g this link).
Concerning observables, the point is that it is quite difficult to give a satisfactory algebraic setting in order to collect together observables that are unbounded (as they actually are the majority of physically relevant quantities: e.g. energy, momentum...). One option is to construct an algebra of unbounded operators, but there are all kinds of domain "nightmares" to be taken into account. Another is to consider an algebra of bounded operators (a $C^*$ or von Neumann algebra), and "affiliate" unbounded self-adjoint operators to it in a suitable fashion. Both procedures are not, in my opinion, completely satisfactory; anyways the algebraic approach gives a very nice framework to understand some of the aspects of quantum theories, especially representations of groups of operators.
My personal point of view is to consider any self-adjoint operator on a given Hilbert space (usually a suitable representation of a $C^*$ algebra, or its bicommutant) as an observable. This choice is justified from the fact that any real-valued physically measurable quantity that is actually measured by physicists behaves like a self-adjoint operator (and not a symmetric one); in particular it has (mathematically speaking) an associated spectral family, as it is the case for self-adjoint operators but not for symmetric ones.
A last mathematical comment: Of course you can associate to a given self-adjoint operator $A$ a strongly continuous unitary group $e^{itA}$; and for example construct the $C^*$ algebra $\{e^{itA},t\in\mathbb{R}\}\overline{\phantom{ii}}$; where the bar stands for the closure (in the operator norm). That algebra may be very interesting to study, and be related to a certain symmetry group of transformations and so on. However, there are other algebras that could be even more interesting, for example the resolvent algebra $\{(A-i\lambda)^{-1}, \lambda\in\mathbb{R}\}\overline{\phantom{ii}}$.
In the case of CCR, the resolvent algebra has a "richer" structure of affiliated self-adjoint operators, and more importantly of automorphisms. That means that more types of quantum dynamics can be defined on the resolvent algebra, preserving it, than for the Weyl (unitary exponential) algebra. In the viewpoint of observables being only the operators affiliated to a given algebra, this means that the resolvent algebra contains more observables, and a less trivial structure of possible evolutions than the Weyl algebra.
Best Answer
One problem with the given $3\times 3$ matrix example is that the eigenspaces are not orthogonal.
Thus it doesn't make sense to say that one has with 100% certainty measured the system to be in some eigenspace but not in the others, because there may be a non-zero overlap to a different eigenspace.
One may prove$^{1}$ that an operator is Hermitian if and only if it is diagonalizable in an orthonormal basis with real eigenvalues. See also this Phys.SE post.
$^{1}$We will ignore subtleties with unbounded operators, domains, selfadjoint extensions, etc., in this answer.