Your interpretation is not quite right. One sharp interpretation one can give to this "cutting" of phase space into cubes of size $h^{2N}$ (here $N$ is the dimension of the system's configuration space), is that it allows one to use classical phase space to count the number of energy eigenstates of the corresponding quantum hamiltonian. Instead of trying to describe what I mean, let's investigate this stuff through an example.
Consider, the one-dimensional simple harmonic oscillator. The hamiltonian is
$$
H(q,p) = \frac{1}{2m}p^2 + \frac{1}{2}m\omega^2 q^2
$$
Let's say I want to answer the following:
Question. Given an energy $E>0$, how many states are there with energies less than $E$?
Now, we have to be careful here because the term "state" means different things in the classical and quantum cases. In the classical case, a state is a point $(q,p)$ in phase space. In the quantum case, a state is a vector in Hilbert space. We can therefore reinterpret the question as follows:
Classical version. What is the area $A(E)$ of the region of phase corresponding to all classical states $(q,p)$ with energies less than $E$?
Quantum version. How many energies eigenstates $\Omega(E)$ does the Hamiltonian posses with energies less than $E$?
The amazing thing is that provided we measure area in phase space in units of $h$, and provided we consider $h$ to be much smaller than the other scales in the problem, both of these questions will give (approximately) the same answer! Let's show this. In the classical case, the region of phase space containing all states with energies less than $E$ is the area of the interior of the ellipse defined by
$$
E< H(q,p)
$$
It turns out that the area of this ellipse is
$$
A(E) = \frac{2\pi E}{\omega}
$$
On the other hand, in the quantum case recall that the energy eigenvalues are $E_n = (n+1/2)\hbar\omega$. This means that the number of eigenstates having energy less than $E$ is found by solving
$$
(\Omega(E) + \tfrac{1}{2})\hbar\omega =E
$$
which, for $h$ small gives
$$
\Omega(E) \sim \frac{E}{\hbar\omega}
$$
Now here's where the magic happens, notice that
$$
\frac{A(E)}{\Omega(E)} \sim \frac{\frac{2\pi E}{\omega}}{\frac{E}{\hbar\omega}} = 2\pi \hbar = h
$$
so we have
$$
\boxed{\Omega(E) \sim\frac{A(E)}{h}}
$$
In words: the area of phase space, measured in units of $h$, allows us to accurately count the number of quantum states below a given energy
The strongest limit without loss of generality is
$$
\Delta p\Delta x \ge \frac12 \hbar,
$$
this is always true. Whilst $\Delta p\Delta x \ge \hbar$ might often be true, it is not always true.
The $\frac12$ is often omitted, because, as mentioned in the comments, often only the magnitude of the right-hand-side is important, and not its precise value. Also, it might be omitted for brevity/simplicity.
A further reason is historical: Heisenberg's original statement of his uncertainty principle was a rough estimate that omitted $\frac12$. Only later was his estimate refined with a formal calculation and the $\frac12$ added.
Best Answer
If you take a particle in a box (3D infinite square well), then the density of momentum states per unit volume is given by $$ g(p)\ dp = g_s\ \frac{4\pi p^2}{h^3}\ dp,$$ where $g_s$ is a spin degeneracy factor (e.g. $g_s=2$ for electrons, protons etc. because they can have 2 spin states per momentum state). This means that the volume of phase space occupied by one momentum state must be $h^3$.
The treatment you give in your question is a rough attempt to justify this result in terms of the (usually) more familiar Heisenberg uncertainty principle. Such a treatment cannot be exact since for a particle in a 1D infinite potential well, the product of uncertainty in momentum and position is actually given by $$ \Delta p \Delta x = \frac{h}{4\pi} \left(\frac{n^2 \pi^2}{3} - 2\right)^{1/2}$$ (see this link), where $n \geq 1$ is the principle quantum number and thus the product of momentum and position uncertainty is always bigger than $h/4\pi$ and would then need to be averaged in some way over all the momentum states.