In Lagrange's formulation we know that $q,\dot {q}$ are independent of each other i.e,
$$\frac { \partial q }{ \partial \dot { q } } =0.$$
My question is, is this true for $p$, $q$ in Hamilton's formulation? Is it true that,
$$\frac { \partial p }{ \partial \dot { p } } =0, \qquad \frac { \partial q }{ \partial \dot { q } } =0~?$$
In Lagrangian formulation, $q$ and $\dot{q}$ are (independent) coordinates and the latter becomes the temporal derivative of the former just on the curves solving the equations of motion. Indeed Euler-Lagrange equations read $$\frac{d}{dt}\frac{\partial L(t, q(t), \dot{q}(t))}{\partial \dot{q}}-\frac{\partial L(t,q(t), \dot{q}(t))}{\partial q}=0\:, \quad \frac{dq}{dt}= \dot{q}(t)\:.$$
In Hamiltonian formulation, $q$ and $p$ are the (independent) coordinates and Hamilton equations read $$\frac{dp}{dt}= - \frac{\partial H(t,q(t),p(t))}{\partial q}\:, \quad \frac{dq}{dt}= \frac{\partial H(t,q(t),p(t))}{\partial p}\:.$$
Here $\dot{q}$ and $\dot{p}$ are not defined coordinates, so the question does not make much sense.
We assume that OP already understands why $q$ and $\dot{q}$ are independent variables in the Lagrangian $L(q,\dot{q},t)$ (which was the topic of this Phys.SE post), and jump right into the Hamiltonian formalism.
The short answer is that the Hamiltonian $H(q,p,t)$ only depends on phase space variables $q$, $p$, and time $t$. It is not supposed to depend on time derivatives $$\dot{q}, \dot{p}, \ddot{q}, \ddot{p},\ldots .$$
If the Hamiltonian $H(q,p,t)$ does depend on time derivatives, something most likely went wrong during the Legendre transformation from the original Lagrangian formulation.
By the way, for a Lagrangian $L(q,\dot{q},\ddot{q},\ldots,t)$ with higher time derivatives, one may wonder how the corresponding Hamiltonian formulation avoids time derivatives? Well, Ostrogradsky told us how: Introduce more canonical pairs!