what physical quantity do real scalar field operators create/destroy? Let $\phi(\textbf{x}) \neq \phi^\dagger(\textbf{x})$ be a complex scalar field, and let $\varphi(\textbf{x}) = \varphi^\dagger(\textbf{x})$ be a real scalar field.
$\phi(\textbf{x})$ destroys a particle and/or creates an antiparticle at $\textbf{x}$.
Equivalently, $\phi(\textbf{x})$ destroys a single charge quantum at $\textbf{x}$, while $\phi^\dagger(\textbf{x})$ creates it. Reinterpreting this in terms of "particles" and "antiparticles" gives the same results as above because to create a charge I could equally well create a particle with that charge or destroy the corresponding antiparticle.
This is perfectly compatible with the fact that complex/non-hermitian fields represent charged particles, and from this reasoning we can expect that a chargeless particle must be represented by a real/hermitian field $\varphi$, as in this case there is no charge to create or destroy hence no way to distinguish $\varphi$ from $\varphi^\dagger$, which are thus the same thing (which is related to the lack of $U(1)$ gauge simmetry in the real case, and that of the related conserved charge). These reasonings raise a few questions:


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*The fact that $\phi$ represents creation and destruction of a particle can be naturally interpreted saying that $\phi$ destroys a charge quantum. What does the operator $\varphi$ represents then? What quantity does $\varphi$ destroy that is equivalently achieved by destruction and creation of a chargeless "particle"?

*Consider the Fourier expansion of $\varphi$ (neglecting constants ecc.):
$$ \varphi(\textbf{x}) = \sum_{\textbf{k}} \left( e^{i\textbf{k} \cdot \textbf{x}} a_\textbf{k} + e^{-i\textbf{k} \cdot \textbf{x}} a_\textbf{k}^\dagger \right)$$
For this to be hermitian we must have $a_{-\textbf{k}} = a_\textbf{k}^\dagger$, which I can interpret saying that "adding a momentum quantum $\textbf{k}$" is equivalent to "removing a momentum quantum $-\textbf{k}$". This makes me thing that the $a_\textbf{k}$ operators do not represent the "creationg of a particle with momentum $\textbf{k}$" but rather the addition of a momentum quantum $\textbf{k}$ to the field.
However, I can think of a situation in which I have 2 chargeless particles each one with momentum $\textbf{k}$ (we are talking of bosons so this should be acceptable), and this reasoning leads me to the conclusion that this situation would be equivalent to a single particle with momentum $2\textbf{k}$, which is not what I would expect. I suspect that this is due to the fact that the particle interpretation here is misleading, and we should just talk of "field excitations" throught creation/destruction of quanta of physically measurable quantities. I cannot measure "a particle" but rather its momentum (or whatever), so I should just talk of momentum quanta and forget the "particles". Are this reasonings correct? (SPOILERS: NO)
 A: *

*The field theory is fully analogous for Hermitian and non-Hermitian fields
The Hermitian operator $\varphi$ still creates and/or annihilates particles and the number of these particles $N$ is still well-defined (at least if we ignore interactions and problems with loops and divergences).
The only difference from the non-Hermitian field is that the particles and the antiparticles are the same things, so instead of two different "counts" of a particle species – one for an electron and one for a positron, for example – we only have one species (e.g. the Higgs boson) and one operator $N$.
If the operator $\varphi$ acts on an $N$ eigenstate with the eigenvalue $n$, it produces a superposition of two states, one of which is an $N$ eigenstate with the eigenvalue $n+1$ and the other one has the eigenvalue $n-1$.
In the charged, non-Hermitian case, there is usually a simple quantity that is conserved, the "charge" – it's the difference of the number of positrons minus the number of electrons, for example. In the case of the Hermitian field, there is no analogous operator (except for the discrete $(-1)^N$) that is conserved. But otherwise there is no difference between the two cases.


*It is not true that the modes with opposite momenta are linked
Your conclusion that $a_k=a^\dagger_{-k}$ is simply incorrect. If you rewrite the sum you wrote as the sum of two sums, then the second sum is self-evidently the Hermitian conjugate of the first sum (I didn't want to talk about the terms for a fixed value of $\vec k$, to avoid discussions about the difference between $\vec k$ and $-\vec k$), so the total expression for $\varphi(x)$ is Hermitian without any additional conditions just like $z+\bar z$ is real without any additional conditions on a complex $z$. 
Obviously, it can't be true that a creation operator is equal to an annihilation operator. They are qualitatively different (each annihilation operator annihilates the vacuum, while no creation operator does so) so they cannot be equal.
