There will be vibrations due to the sudden stopping of the system. Whether or not it will "cause great vibration" is really your discretion.
To understand what is happening, the motor shaft has some inertia. The angular momentum of the shaft is denoted by I * w, where "I" is the first moment of inertia of the shaft (in kg*m^2), and "w" (omega) is the angular velocity (in rad/sec). The torque generated on the motor is analogous to the force generated when a moving object suddenly stops. Thus, T = dL/dt, where "T" is torque (in N*m), and "dL/dt" is the change in angular velocity (L) per unit time. Obviously, you can't actually stop instantly, since that would require infinite torque, but you can stop rather quickly.
If you want to actually solve for the dynamics of the system, you need to understand Linear, Time Invariant (LTI) Second-Order Systems. Essentially, you can analyze your stepper motor to determine it's inertia (2nd-order term), damping (1st-order term), and springiness (0th-order term), then use the equation:
I * theta'' + b * theta' + k * theta = T = dL/dt
In this equation, I is your moment of inertia, b is your damping, and k is your springiness. Theta (and it's time derivatives) represent your angle. You could use a solver (like Mathematica/WolframAlpha or MATLAB/Octave) to solve the system given your initial conditions.
Of course, since both "b" and "k" are likely to be small, your system is really more like:
I * theta'' = dL/dt
which is far easier to solve.
If you read up more on this, you could simulate your braking system so that you can see the oscillations if you stop immediately, and find a dL/dt (or braking torque) that creates the time-optimal decay in velocity.
If you want to learn more, see these links: