distinguishable particles' Hamiltonian Let us consider a classical Hamiltonian of a many body system
\begin{equation*}
H = \sum_{j=1}^N\frac{p_j^2}{2m}+V(\mathbf q)
\end{equation*}
and let us pass to quantum dynamics by promoting the positions and momenta to operators:
\begin{equation*}
q_j\rightarrow \hat q_j \;\;\; p_j\rightarrow \hat p_j
\end{equation*}
where the index label still refers to THE particle $j$.
Additionally, we could define a bosonic Fock space via the creation and destruction operators (by taking $\hbar\equiv 1$)
\begin{equation*}
a^\dagger_j=\frac{\hat q_j+\hat p_j}{\sqrt 2} \;\;\;
a_j = \frac {\hat q_j+i\hat p_j}{\sqrt 2}\hspace{8mm} j=1\cdots N
\end{equation*}
creating/destroyting A particle in A STATE $j$. To me the fact that particles are seen as indistinguishable once a Fock space is build appears natural. In particular the label $j$ does not refer anymore to a particle but rather to a state. Instead, if we stick to the quantum dynamical system $\hat H(\mathbf{\hat q}, \mathbf{\hat p})$, is it true that particles are still distinguishable? In particular, I think that in the calculation of the canonical partition function
\begin{equation*}
Z(\beta) = \int_{\mathbb R^N} \mathrm d \mathbf q \langle \mathbf q|e^{-\beta \hat H(\mathbf {\hat q}, \mathbf {\hat p})}|\mathbf q\rangle
\end{equation*} 
the states in the braket doesn't need to be symmetrized (as we would do for a trace in the Fock space) and that we can stick to the tensor product of the single particle Hilbert spaces:
\begin{equation*}
|\mathbf q \rangle = |q_1\rangle \otimes\cdots \otimes |q_N\rangle
\end{equation*}
Is my line of arguments correct?
 A: This seems to be a rather confused question. Given a general interaction potential $V(\mathbf{q}_i)$, your operators $a_j^\dagger$ and $a_j$ simply aren't creation or annihilation operators, in the naive sense. Their commutators with the Hamiltonian are much more complicated.
Assuming that $V(\mathbf{q}_i)$ has the very special form
$$V(\mathbf{q}_i) = \frac12 \sum_i k_i q_i^2$$
then you indeed do get a set of $N$ independent harmonic oscillators. In this case, your conclusions are still confused because you're describing the same state in two incompatible ways. 
For simplicity, let's restrict to $N = 3$ and consider the state
$$|2, 1, 0 \rangle = a_1^\dagger a_1^\dagger a_2^\dagger |0 \rangle.$$
We can describe this state in two ways: either "there are three harmonic oscillators; the first harmonic oscillator is in the second excited state, and..." or "there are three units of excitation; two are in the first harmonic oscillator, and...". 
The source of your confusion is that you're calling the harmonic oscillators themselves the particles in the first case, but then switching to calling the units of excitation the particles in the second case. Of course, "particle" is just a word which you can freely attach to either concept, with one or the other being more useful depending on the situation, but it shouldn't surprise you that the properties of the particles change if you change what the word "particle" means halfway through. 

To hopefully clear some of the confusion, let's do an explicit example. Let $N = 3$ and consider the state $a_1^\dagger a_3^\dagger |0 \rangle$. 


*

*In the original picture, the Hilbert space is $\mathcal{H}_1^{\otimes 3}$ where the tensor product is not symmetrized. We describe this state in words as "there are three particles; the first particle is in the first excited state, the second is in the ground state, and the third is in the first excited state".

*Next, we may switch to a completely different description of the states. We rename "particles" to "modes" and rename "particle in $n^{\text{th}}$ excited state" to "$n$ particles in the mode". Hence to describe the state we say "there are two particles; the first particle is in the first mode, and the second is in the third mode". We can write this state as $|1, 3 \rangle$ or as $|3, 1 \rangle$.

*Since the "particles" are identical, the states $|1, 3 \rangle$ and $|3, 1 \rangle$ are both the same state, namely $a_1^\dagger a_3^\dagger |0 \rangle$. So we must work with a symmetrized Fock space.


To say the distinction yet another way: suppose I own $3$ sheep among a large herd, and to pick them out I give them identical bells. In the first langauge, the particles are the sheep, and I specify the state by saying where they are. In the second language, the particles are the bells, and I specify the state by saying what sheep each bell is on. Since the bells are identical, I then have to quotient out by permutations of the bells. But that doesn't mean the sheep have become identical.
