It appears that stationary states aren't so stationary $$
\DeclareMathOperator{\dif}{d \!}
\newcommand{\ramuno}{\mathrm{i}}
\newcommand{\exponent}{\mathrm{e}}
\newcommand{\ket}[1]{|{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}|}
\newcommand{\braket}[2]{\langle{#1}|{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}|{#2}|{#3}\rangle}
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\dpd}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\dod}[2]{\frac{\dif{#1}}{\dif{#2}}}
$$
Using the Schrödinger equation and the definition of the expectation value it can be shown that the time dependence of the expectation value of an observable $A$ for a system in arbitrary state $\ket{\Psi(t)}$ is given by
$$
    \dod{\langle A \rangle}{t} = \frac{\ramuno}{\hbar} \langle [\linop{H}, \linop{A}] \rangle + \big\langle \dpd{\linop{A}}{t} \big\rangle \, , \tag{1}
$$
and this equation shows that, in general, if an operator $\linop{A}$ commutes with the Hamiltonian operator $\linop{H}$ and does not have an explicit time dependence, then the expectation value of the corresponding observable $A$ is time independent.
For stationary states $\ket{\Psi(t)} = \exponent^{-\ramuno E_{k} t / \hbar} \ket{E_{k}}$
the first term in the expression for the time dependence of the expectation value of an observable vanishes
$$
    \langle [\linop{H}, \linop{A}] \rangle = 
    \bracket{ \Psi(t) }{ \linop{H} \linop{A} }{ \Psi(t) } - \bracket{ \Psi(t) }{ \linop{A} \linop{H} }{ \Psi(t) } =
    \bracket{ E_{k} }{ \linop{H} \linop{A} }{ E_{k} } - \bracket{ E_{k} }{ \linop{A} \linop{H} }{ E_{k} } = 
    E_{k} \bracket{ E_{k} }{ \linop{A} }{ E_{k} } - E_{k} \bracket{ E_{k} }{ \linop{A} }{ E_{k} } = 0 \, ,
$$
and so the time dependence of the expectation value is simply given by
$$
    \dod{\langle A \rangle}{t} = \big\langle \dpd{\linop{A}}{t} \big\rangle \, . \tag{2}
$$
Nevertheless, the statement like the following one 

A stationary state is called stationary because the system remains in the same state as time elapses, in every observable way. Wikipedia

is found in many books.
The thing that troubled me is the word every, since from (2) it appears that if an operator $\linop{A}$ carries some explicit time dependence, then the expectation value of the corresponding observable $A$ changes in time.
So stationary states are, in fact, not so stationary. 
I have the feeling that I am missing something.
And our discussion with Bubble helped clarify what's bothering me.
As far as I know operators in the Schrödinger picture usually do not carry an explicit time dependence. Again, usually, but not always.
In many books (see, for instance, Griffiths, D.J., Introduction to quantum mechanics, 2nd ed.) one can find that

Operators that depend explicitly on $t$ are quite rare, so almost always $\dpd{Q}{t} = 0$. 

And, yet the author claims that

Every expectation value is constant in time.

I feel like there is a gap between operators being almost always explicitly independent of time and every expectation value being constant.
 A: I think it's expected that you have a bit of common sense about this.
Let's take the operator $O(t)$ which is the position operator when $t$ is between 9:00 and 10:00 in the morning, and the momentum operator the rest of the day.
Now take a system in a stationary state, and ask "What is the expectation value of $O(t)$ at each time $t$"? Whoa, the expectation value changes dramatically each day at 9:00, then changes again at 10:00!
Does that mean the state is not "stationary" at 9:00 or 10:00? No, of course it doesn't mean that!
When an operator has explicit time dependence, as $O(t)$ does, it means that you have a different -- possibly totally unrelated -- operator at each time $t$. Wikipedia says "the system remains in the same state as time elapses, in every observable way". That's correct. I don't think a reasonable person reading that sentence would infer that if you calculate the expected position at 9:30, and then you calculate the expected momentum at 10:30, the two calculations should have the same answer for a stationary state.
A: I'm not really sure anymore what you're asking so I will try to answer all possible variations of your question that I can think of. 


*

*If you are asking whether there exist observables which are explicitly time dependent while the Hamiltonian of your system is time independent then the answer is yes. For instance, let's say that you can define momentum,$\hat{p}$, for your system then the operator $\hat{O}=t \hat{p}$, which is linear and Hermitian, is an observable. You can measure it somehow. For instance, you can measure momentum and then multiply it by the time you see on your clock. After measurement the system will be in one of the momentum eigenstates, of course.

*If you are asking can a Hamiltonian in the Schrödinger picture evolve an operator so that the operator is time dependent (either explicitly or implicitly) then the answer is no. All of the time dependence is in the state vector by definition. 
You can put a field which varies with time, for instance, in the Hamiltonian, but then you can not use $|\psi(t)>=e^{−iE_kt}|E_k⟩$ anymore. Note that the article assumes that you have a time independent Hamiltonian. You have to use Dyson series to solve the equation of motion. And in any case, if you are in the Schrödinger picture the time dependence will again be carried by the state vector, except for the operators you defined to have explicit time dependence. 
A: Imagine a ball, laying still on the ground.  While you are looking at it, slowly walking around. The state of the system is staying the same, while the things that you see depends on your aim.
State of the system -- only it matters. And that is what your wiki quote clearly utters.  


Looking for an answer drawing from credible and/or official sources.

Edit: All this discussion is one big nitpicking. Remark you are making is not really sticking. For "official sources" it won't worth the fuss. If you want clarity, then just use the math. 
