Hamilton equations from Poisson bracket's formulation
The confusion stems from a bit of mathematical sloppiness. It's necessary useful sloppiness, because as you'll see in a moment, the full machinery is a pain, but it's helpful to see and keep in the back of your mind when confusions like this arise.
Phase Space Trajectories
I won't go through the trouble of constructing the cotangent bundle to the configuration space and all of that mess - we can start from the intuitive notion of the phase space $\Omega$. A point $x\in\Omega$ can be labeled by the corresponding position $q$ and momentum $p$ (i.e. $x\equiv(q_x,p_x)$). It's important to note that $q$ and $p$ are not functions of time or anything else - they are just numbers which label a particular location in phase space.
From here, we consider the notion of a trajectory through phase space. A trajectory $\gamma$ is a continuous map which takes a real number (the time) and maps it to a point in phase space:
$$\gamma:\mathbb{R} \rightarrow \Omega$$ $$t \mapsto \gamma(t)$$
If we feed $\gamma$ a time, it tells us the location of the system in phase space. As we move forward in time, the trajectory tells us how the state of the system evolves.
Dynamical Variables
A dynamical variable $F$ takes a point in phase space and a value of time and maps them to a real number: $$F :\Omega\times \mathbb{R} \rightarrow \mathbb{R}$$ $$(q,p,t) \mapsto F(q,p,t)$$
Next we introduce the projection functions $\mathcal{Q}$ and $\mathcal{P}$, which map a particular point in phase space to the corresponding values of $q$ and $p$. $$\mathcal{Q}:\Omega\times\mathbb R \rightarrow \mathbb{R} $$ $$(x,t)\mapsto \mathcal{Q}(x,t)\equiv\mathcal{Q}\big(q_x,p_x,t\big) = q_x$$ and $$\mathcal{P}:\Omega\times\mathbb R \rightarrow \mathbb{R} $$ $$(x,t) \mapsto \mathcal{P}(x,t)\equiv\mathcal{P}\big(q_x,p_x,t\big) = p_x$$
Essentially, $\mathcal{Q}$ just takes a point in phase space and a time and tells you the position coordinate while ignoring the momentum coordinate and the time, while $\mathcal{P}$ takes a point in phase space and a time and tells you the momentum while ignoring the position coordinate and the time. Notice that these two functions are particular examples of time-independent dynamical variables, in the sense that $\frac{\partial \mathcal{P}}{\partial t} = \frac{\partial \mathcal{Q}}{\partial t}=0$.
Dynamical Variables Along Phase Space Trajectories
Now we can combine these two concepts. Given a trajectory $\gamma$ and a dynamical variable $F$, we can combine them to form a map $F_\gamma$ which takes a single real number $t$ and returns the value of $F$ at time $t$ along $\gamma$:
$$F_\gamma : \mathbb{R} \rightarrow \mathbb{R}$$ $$ t \mapsto F\big(\gamma(t),t\big)$$
We can apply this definition to $\mathcal{Q}$ and $\mathcal{P}$. Notice that $$\mathcal{Q}_\gamma : \mathbb{R} \rightarrow \mathbb{R}$$ $$ t \mapsto \mathcal{Q}\big(\gamma(t)\big)$$ so $\mathcal{Q}_\gamma(t)$ is the position coordinate of the system at time $t$, while $\mathcal{P}_\gamma(t)$ is the momentum coordinate of the system at time $t$.
Notice that for a given dynamical variable $F$, we can also write that $$F_\gamma(t) = F\big(\gamma(t),t\big) = F\big(\mathcal{Q}_\gamma(t),\mathcal{P}_\gamma(t),t\big)$$
Total Derivatives
Because such maps are functions of a single variable, it makes sense to take a total derivative with respect to time. This is the total rate of change of $F$ along the trajectory $\gamma$:
$$\frac{dF_\gamma}{dt} \equiv \frac{\partial F}{\partial q}\frac{d\mathcal{Q}_\gamma}{dt} + \frac{\partial F}{\partial p}\frac{d\mathcal{P}_\gamma}{dt} + \frac{\partial F}{\partial t}$$
Hamilton's Equations
Hamilton's equations are the differential equations which govern phase space trajectories. Without delving into their derivation, they tell us that $$ \frac{d\gamma}{dt} \equiv \left(\frac{d\mathcal{Q}_\gamma}{dt},\frac{d\mathcal{P}_\gamma}{dt}\right) = \left(\frac{\partial\mathcal{H}}{\partial p},-\frac{\partial\mathcal{H}}{\partial q}\right)$$ where $\mathcal{H}$ is the Hamiltonian - yet another dynamical variable.
Once the Hamiltonian $$\mathcal{H}:\Omega \times \mathbb{R} \rightarrow \mathbb{R}$$ $$(q,p,t) \mapsto \mathcal{H}(q,p,t)$$ has been written down, then all possible phase space trajectories have been determined.
Poisson Bracket
Using Hamilton's equations, we can rewrite the total derivative in the following way: $$ \frac{dF_\gamma}{dt} = \frac{\partial F}{\partial q}\frac{d\mathcal{Q}_\gamma}{dt} + \frac{\partial F}{\partial p}\frac{d\mathcal{P}_\gamma}{dt} + \frac{\partial F}{\partial t} = \left(\frac{\partial F}{\partial q}\frac{\partial \mathcal{H}}{\partial p} - \frac{\partial F}{\partial p}\frac{\partial \mathcal{H}}{\partial q}\right) + \frac{\partial F}{\partial t}$$
This motivates the definition of the Poisson bracket of two dynamical variables: $$ \{A,B\} \equiv \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p} \frac{\partial B}{\partial q}$$
at which point we can rewrite the total derivative formula one last time:
$$\frac{dF_\gamma}{dt} = \{F,H\} + \frac{\partial F}{\partial t}$$
All done! Notice that the right hand side makes no mention of the trajectory $\gamma$, for good reason - once we specify the Hamiltonian and our particular location in phase space, then there's no freedom left in the evolution of the system (and therefore no freedom left in the evolution of any dynamical variable).
The Punchline
We're now equipped to answer your question. Consider the function $\mathcal{Q}$ (the function which takes a phase space point $(q,p)$ and returns its position coordinate $q$), as well as the associated function $\mathcal{Q}_\gamma$ which is associated to a phase space trajectory. We have that
$$\mathcal{Q}(q,p,t) = q$$ so $$\frac{\partial \mathcal{Q}}{\partial q} = 1$$ $$\frac{\partial \mathcal{Q}}{\partial p} = 0$$ $$\frac{\partial \mathcal{Q}}{\partial t} = 0$$
and therefore
$$\frac{d\mathcal{Q}_\gamma}{dt} = \{\mathcal{Q},\mathcal{H}\}+\frac{\partial \mathcal{Q}}{\partial t}$$ $$ = \left(\frac{\partial \mathcal{Q}}{\partial q}\frac{\partial \mathcal{H}}{\partial p} - \frac{\partial \mathcal{Q}}{\partial p}\frac{\partial \mathcal{H}}{\partial q}\right) + \frac{\partial \mathcal{Q}}{\partial t} $$ $$= \frac{\partial \mathcal{H}}{\partial p}$$
and similarly, $$\frac{d\mathcal{P}_\gamma}{dt} = -\frac{\partial \mathcal{H}}{\partial q}$$
So there you have it. When we write everything out in excruciating detail, there is no ambiguity whatsoever. $q$ and $p$ are numbers, not functions, so differentiating them is meaningless. When we do physics, what we're actually differentating are the projection functions $\mathcal{Q}$ and $\mathcal{P}$, as well as their associated functions which are attached to the phase space trajectory $\gamma$ along which the system evolves. Again, it's crucial to notice that $\mathcal{Q}$ and $\mathcal{Q}_\gamma$ are associated with each other, but are not the same thing.
Of course, when I do physics, I don't write all this out - I differentiate $q$ and $p$ just like everybody else. But it's useful to be able to frame problems in this context when those little points of confusion arise.
The "partial" time derivative $\frac{\partial}{\partial t}$ means in this context an explicit time differentiation. A function $$(q,p,t)~\mapsto~ f(q,p,t)$$ of phase space and time is said to have an explicit time dependence through its last argument $t$ and an implicit time dependence through the phase space variables $q^i$ and $p_j$. The total time derivative $df/dt$ is then given by OP's eq. (1).
Short explanation: To make OP's eq. (1) also work for the phase space variables $$f(q,p,t)=q^i\quad\text{and}\quad f(q,p,t)=p_j$$ themselves, it is natural (and in practice useful) to pragmatically declare that they depend by definition only implicitly on time.
Longer explanation: Note that physicists (unlike mathematicians) often use the same notation for a function and its value at a point. Therefore it may sometimes becomes difficult to know the list of arguments of a function. In the present case, the notion of phase space variables $q^i$ and $p_j$ can have different list of arguments $$q^i(q,p,t)=q^i\quad\text{and}\quad p_j(q,p,t)=p_j$$ versus $$t~\mapsto~ q^i(t)\quad\text{and}\quad t~\mapsto~ p_j(t)$$ depending on context. Implicit and explicit dependence are strictly speaking only defined in the former case.
See also this Phys.SE post and links therein.
I do not agree with the last part of the answer by Qmechanic. From a mathematical standpoint, the phase space variables depend explicitely on time, as time (as a parameter for curves in phase space) is their only functional variable. If we had, let's say $q^{i} (t) = q^{i} (z(t))$, then yes, the time dependence would have been implicit (i.e. through another function $z(t)$).
Coming back to the question in the OP. The functions $q^i (t)$ and $p_i (t)$ have only one independent variable, namely time $t$. The equation of motion for a generic observable on phase space $F (q,p,t)$ does not apply to them, because the time dependence of these functions $q$ and $p$, compared to the one for the observable, is NOT both implicit and explicit, it is only explicit. This can be reworded to say that $q^i (q^i(t),p_i (t), t)$ is not a valid F. It doesn't even make sense mathematically (similarly $p_i (q^i(t),p_i (t), t)$).