In pages 128-129 of his book "Basic Algebra I", 2nd edition, Dr. Nathan Jacobson proves that the quotient and the remainder exist and are unique as long as the coefficient of the leading term of the divisor is a unit.
In page 114 of his book "Advanced Modern Algebra", 2nd edition, Dr. Joseph Rotman asks (in exercise 2.51(ii)) for an example of a commutative ring with unity where polynomial division by a monic divisor does not results in a unique remainder.
Despite the proof of Jacobson, I tried to find such a division in the ring $\mathbb Z_4[x]$, but I could not succeed.
Dr. Rotman has an example 6.12, page 238, in his book "Learning Modern Algebra", in the above ring $\mathbb Z_4[x]$, but the divisor is not a monic polynomial:
$$2x^3+3=(x+1)(2x^2+2x+1)+(x+2)=(x+3)(2x^2+2x+1)+x$$
What do I miss? (Who is right and who is wrong?)
Well, Rotman is the one who is wrong. You can prove it yourself, if $qq_1+r_1=p=qq_2+r_2$ with $\deg r_i<\deg q$ and $q$ monic, then $$q(q_1-q_2)+(r_1-r_2)=0.$$
If $q_1\neq q_2$, then $q_1-q_2$ has finite degree (is non-zero). Since $q$ is monic, the term of largest degree of $q$ multiplied by the term of largest degree of $q_1-q_2$ gives you the term of largest degree of $q(q_1-q_2)$ because its coefficient cannot be zero if it wasn't already zero the coefficient of the term of largest degree of $q_1-q_2$, which it isn't by definition. The polynomial $r_1-r_2$ has degree smaller than the degree of $q$, therefore it won't cancel the term of largest degree of $q(q_1-q_2)$. This contradicts that $q(q_1-q_2)+(r_1-r_2)=0$. Therefore, $q_1=q_2$. Now, if $q_1-q_2=0$ it follows that $r_1-r_2=0$.