Note: ChocoPouce's answer is the same as mine but is more mathematical.
You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this distribution is in at the $0.1\mathrm{nm}$ shell? We won't exactly $0.1\mathrm{nm}$, so we take a thin shell: how much is between $0.1$ and $0.1+ε\mathrm{nm}$? Amount $=$ (average prob. density in shell)*(volume of shell). Since the shell is so thin ($\varepsilon$ is very small), the density will be almost constant and the shell's volume is given by $\mathrm{area}\times\mathrm{thickness} = 4\pi*(0.1nm^2)*\varepsilon nm$.
Thus the probability is $4\pi\rho(r)r^2\varepsilon$, where $\rho$ only depends on $r$ since it is spherically symmetric. But $\varepsilon$ is an arbitrary "small" thickness we defined, and it is best to divide by $\varepsilon$ to get the probability per unit radius, which is called the (radial) probability density $4\pi\rho(r)r^2$. The most likely radius (which is different from the average radius) maximizes this function.
In QM, a particle is described by its wavefunction. Each state of the hydrogen atom has a particular wavefunction and a particular energy. Since you can count the number of different wavefunction, the energy is quantized.
It is important to understand that the wavefunction does not tell you where the electron is, it can only tell the probability to find the electron in a certain region. Thus, if you try to locate the electron, you will always get different results. But, if you measure the average distance to the nucleus, your result will approach the bohr radius.
Because of this, it looks like the electron is present everywhere in space because the wavefunction is defined everywhere.
The wavefunction also contains information about the energy of the system. If the electron is in the ground state, its energy will be definite (the Rydberg energy). Since you do not know where the electron is exactly, you cannot compute its energy classicaly. If you try to measure the location of the particle and compute the energy from the position you get, you will not know its kinetic energy because of the Heisenberg uncertainty principle ($\Delta x\Delta p \geq\hbar$). Indeed, if you locate the particle, the uncertainty of its position will be small, but because of the principle, there will be a great uncertainty on its momentum and thus, its kinetic energy. It is possible to measure the energy of a particle with great precision, but because of this same principle, you won't know exactly where the particle is.
So, to answer your question, the electron's energy is quantized in the hydrogen atom. But, you cannot know the position of the particle and its energy with great precision at the same time.
Best Answer
In a word: No. The Bohr radius is a key concept and it is not deprecated.
In the modern outlook, the Bohr radius is the length unit of the atomic system of units, i.e., it is the natural length scale that comes out as a combination of the reduced Planck constant $\hbar$, the electrostatic interaction constant $\frac{e^2}{4\pi\epsilon_0}$, and the electron mass $m_e$. When doing atomic and molecular physics as well as quantum chemistry, all calculations are done in multiples of the Bohr radius.
And yes, it does not have a simple definition as "the radius of the ground-state orbit" that it does in the (deprecated) Bohr model, but that does not mean it is not useful.