No, at the distances we have sent any spacecraft or probe to, there is no need to compensate for redshift (the Doppler effect) because a spacecraft is far away. Doppler shift is an effect of relative velocity, not distance.
The Doppler effect is defined such that the frequency $f$ as observed by a receiver moving at velocity $v_r$ relative to a source moving at velocity $v_s$ and where the source is transmitting at frequency $f_0$ is $$ f = \left( \frac{c + v_r}{c + v_s} \right) f_0 $$
In the case of radio transmissions, $c$ is the speed of light. $v_r$ and $v_s$ are measured in some inertial (that's basically a fancy word for "non-moving") reference frame.
For a stationary sender or receiver, simply set $v_s = 0$ or $v_r = 0$ respectively. From this it is obvious how, when both sender and receiver are stationary (in an inertial reference frame), the transmitted frequency equals the received frequency.
As you can see, for a given type of transmission (such as a radio transmission), only the relative velocities of the sender and receiver and the transmission frequency affect the received frequency.
This effect is caused by the fact that when the sender is moving toward the receiver, each wave front is closer to its neighboring wave fronts than merely its frequency would warrant. Likewise, when the two are moving away from each other, each wave front is farther away from its neighbors. Colloquially, the transmission is being "pushed together" or "dragged apart" by the relative movement of the sender and receiver. This is observed as a change in frequency at the receiver.
The distance between the two do not play a role in this effect, and thus we do not need to compensate for Doppler shift merely because an object is far away. We do however have to compensate for Doppler shift because the object is moving relative to us. Of course, because spacecraft generally move at relatively high speeds relative to an Earth ground station, in spacecraft communications we often do need to pay attention to Doppler shift, regardless of whether they are close to or far from Earth.
For example, if the spacecraft is moving away from the Sun at 15 km/s (not unreasonable at all for interplanetary missions) and Earth happens to be moving in the opposite direction at 30 km/s (which it does once a year), and the spacecraft is transmitting on 5 GHz, the frequency as received on Earth will be $$ f = \left( \frac{300~000~000 - 15~000}{300~000~000 + 30~000} \right) \times 5~000~000~000 \approx 4~999~250~075 $$ or approximately 4999.25 MHz instead of 5000.00 MHz. Conversely, if we want the spacecraft to receive our transmission on 5000.00 MHz, then we need to transmit at approximately 5000.75 MHz. If we didn't compensate for Doppler shift, we would need a receiver bandpass filter bandwidth of nearly a megahertz; enough to cram in two or three broadcast FM stations.
It's also worth keeping in mind that this is an issue not just for deep space missions. The math is a bit more involved because of the trigonometry to calculate the absolute value of the velocity and if you want an absolute value also compensating for the planet's rotation, but the exact same principles apply when a low orbit spacecraft is communicating with a ground station or for that matter with a spacecraft in a different orbit.
If we were to send spacecraft to other galaxies or even far-away stars, then Doppler shift due to the expansion of the universe might start to actually matter. However, for as long as it can be argued whether we have even exited our own solar system, the relative velocities caused by expansion of the universe are of magnitudes that are completely irrelevant for practical radio communications. The expansion of the universe is given as Hubble's constant $H_0$, and the most recent estimate for its value is $67~800 \pm 770$ m/s per megaparsec. The Voyagers are about 134 and 110 AU away, respectively, which works out to $6.497 \times 10^{-10}$ and $5.333 \times 10^{-10}$ megaparsec. If my math is right, taking Voyager 1 because it is farther away, this means an observed expansion rate of about 0.00004405 m/s (44 µm/s) corresponding to a received frequency of $$ \left( \frac{300~000~000}{300~000~000 + 0.00004405} \right) \times 5~000~000~000 = 4~999~999~999.999265833 $$ or a Doppler shift of about 0.001 Hz when the transmitted frequency is exactly 5 GHz. In other words, valid in theory but completely negligible in practice.