could we also assume that they can alter the signal and therefore the sound?
Well, yes, but normally in exactly the way you have experienced here... either the signal passes or it doesn't.
If you need more reassurance, consider how the digital signal moves through the cable.
One sample of a 16-bit signal looks like this:
1101001010001001
The next sample might be:
1101001010001011
There are 44,100 of these per second moving through the cable.
They are represented by high and low voltages moving through the cable, high voltage =1, low voltage =0. I am oversimplifying it since there is a protocol around the raw data which I frankly don't fully understand, but
this is the basic idea. @myiot was able to provide
better explanation here.
To alter the frequency response of an analog signal, sometimes just a certain value of capacitance in the circuit is enough. But to alter the frequency response of a DIGITAL signal, you need to operate on multiple samples at once, and have a feedback loop that adds/subtracts/multiplies the samples together. This is how digital filters are constructed.
But if there is something wrong with the cable, it will only affect individual 1s and 0s going by. There is not even a way for full samples to be affected in a consistent way, let alone entire groups of samples being mathematically compared to each other.
So if there is digital interference, generally you will only hear an interruption in the whole signal (as you heard here) or random noise. If you randomly "flip bits" in digital signals, the audible result is simply noise. Because of the protocols I mentioned earlier, often there is some protection against noise also. (Consider that your bank transactions or video downloads don't get any "random noise"... ever, and we can see why digital audio is so robust, too. "Bits are bits". )
So don't worry about losing high frequencies or anything like that over a digital cable, that kind of audible effect is essentially impossible.
