Short answer
No, you cannot make water boil simply by spinning it in a glass. It's impossible. The pressure inside the water will not decrease, but increase. Even at the water surface.
Long answer
The situation
Assume there is a glass cylinder, which is closed at the bottom and initially open at the top. It is partially filled with water and spinning at a very high velocity (a very, very, very high velocity if you like). The water is dragged along with the glass cylinder and at some point spins with the same velocity as the glass, generating a vortex.
At some velocity (depending on the length of the glass and the amount of water), the water will start to spill out on top of the glass. To prevent this, we place either a ring on top, leaving a hole in the center, or we close the glass completely. This does not make a big difference.
There are two phases involved: Water and air. Let's first look at ...
The water.
The pressure distribution inside the water can be calculated using e.g. the idea of potential energy. There effectively are two pressure gradients which are superposed:
the usual hydrostatic pressure gradient due to Earth's gravity from top to bottom (roughly 1 bar per 10 m),
a pressure gradient due to the rotation from the center line outwards.
The following formula can be derived for the pressure $p$ inside the water:
$$
p = p_0 - \rho g (z-z_0) + \tfrac 12 \rho \omega^2 r^2
$$
($z$ vertical coordinate pointing upward, $r$ radial coordinate pointing outward, $\rho$ fluid density, $g$ gravity of Earth, $\omega$ angular velocity, $p_0=p(z=z_0,r=0)$).
Note that the assumptions that result in this pressure distribution are only a spinning tube in Earth's gravity field, filled with incompressible fluid. No assumption has been made yet on the position of the boundary between water and air.
This means that the pressure is increasing with increasing $r$. At the boundary between water and air, the pressure is lowest, while at the inner wall of the cylindrical part of the glass, the pressure is highest. Due to Earth's gravity, the highest pressure is reached at the bottom of the glass.
Important statement: The pressure inside the water is increasing when moving from the inside (surface) to the outside (glass).
Now let's look at ...
The air.
For the air that spins inside the cylinder, the same principle holds true as for the water. Since air is compressible, the above equation cannot be used directly for air, but a similar statement can be made:
In the air, too, the pressure increases when we move from the inside (rotation axis) to the outside (water surface).
This by the way also holds true for tornados: The outside pressure is greater than the inside pressure. For tornados, the outside pressure is set by the surrounding atmospheric pressure.
One big difference is of course the density: $\rho_{\mathrm{water}} \approx 1000 \mathrm{kg/m^3}$, $\rho_{\mathrm{air}} \approx 1.2 \mathrm{kg/m^3}$. This results in a much smaller pressure gradient inside the air when compared to the pressure gradient in the water. But for very, very high velocities, the pressure gradient in air can become non-negligible.
Now we need ...
The Euler–Cauchy stress principle.
This is one of the fundamental assumption of continuum mechanics. It states that the stress vector is continuous. For the water and the air in our glass this means that the pressure at the water-air boundary is equal inside the water and the air. Put in other words: The pressure between water and air is continuous and does not jump.
Now we finally need the ...
Reference pressure.
This is of course der springende Punkt, as we say in German ("the jumping point" - the important point).
If the tube is partially open, the pressure will equalize at the opening. Since the tube is spinning really, really fast, there is a pressure gradient inside the air as well. This will result in air flowing into the tube at the rotation axis, and air flowing out of the tube at the outer rim of the opening. So we get a secondary flow here. This secondary flow leads to a decreased pressure through the Bernoulli equation (see below), but the pressure at the opening cannot drop below atmospheric pressure, because otherwise the secondary flow would stop.
For a stationary rotation, this secondary flow is stationary. Because the air pressure near the rotation axis must be smaller than the air pressure at the water surface, air must flow in near the rotation axis and exit the tube near the outer rim of the opening (assuming we do not consume or generate matter inside the tube).
Therefore, the pressure at the outer part of the opening must be above atmospheric pressure.
Since the pressure gradient from rotation axis to water surface extends throughout the whole length of the tube, the secondary flow also extends through the whole length of the tube. In order to keep going, the pressure at the water surface at the bottom of the tube must therefore be slightly above the pressure at the outer rim of the opening.
Therefore, the pressure at the entire water surface must be above atmospheric pressure.
The closed tube is easier. There is no secondary flow, just a pressure increase towards the water. Since the amount of air inside the tube has not changed, and the pressure at the water surface is above the pressure at the tube center, the pressure at the water surface must be above atmospheric pressure. The pressure at the tube center is below atmospheric pressure.
In total, all you can get is a pressure increase at the water-air boundary as @Peter.A.Schneider pointed out in his comment. This is independent on how fast the glass spins.
For an open glass, the secondary flow will by the way result in water being evaporated, which will lead to a cooling of the water. So instead of boiling it is more likely that the water will freeze.
Bernoulli equation
Finally, the original posts states:
I think the pressure in a fluid reduces when the speed increases
A pressure decrease in a moving fluid is relative to the fluid at rest. You are referring to the momentum conservation equation for an inviscid fluids, aka Bernoulli equation:
$$
\tfrac12 (v_2^2-v_1^2) + \frac 1 \rho (p_2-p_1)=0
$$
(here without the gravitation term). This is valid along a streamline only. It is not a general principle that the pressure decreases in fast moving fluids.
In the spinning glass, the streamlines are circles concentric to the spin axis. Along these streamlines, neither velocity nor pressure changes. In short, using Bernoulli's equation on your spinning glass does not tell you anything on the pressure in the water.
