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
The creation and annihilation operators are not observables. They are obviously not hermitean because $a \neq a^\dagger$. But, regarding your question, consider the number operator $( N_k = a_k a^\dagger_k )$. As the eigenvalue of the number operator is a dimensionless number, the creation and annihilation operators must be dimensionless as well.
The full Hamiltonian has the dimension of energy, because $[\hbar] = \mathrm{J s / rad}$ and $[\omega = \mathrm{rad / s}$, so $[\hbar \omega] = \mathrm{J}$.