Three easier warm-up questions and then The Real Question...
Question 1: Do there exist numbers $p$ and $q$ in $\mathbb Q$ such that in order to have $pr=q$ we must have that $r$ is irrational?
Answer to Question 1: No. Either $p=q=0$, in which case $r$ can be anything, or $q\neq0$, in which case $r=\frac qp$ is the only solution, and this is rational.
Question 2: Do there exist polynomials $p(x)$ and $q(x)$ in $\mathbb Q[x]$ such that in order to have $p(x)r(x)=q(x)$ we must have that $r(x)$ has at least one irrational coefficient?
Answer to Question 2: No. Actually, given that $q(x)\in\mathbb Q[x]$ is nonzero, it is not possible for $q(x)$ to be expressed as the product of $p(x)$ and some polynomial $r(x)$ unless $r(x)$ has strictly rational coefficients. This is less obvious than the situation with the previous question.
Question 3: Do there exist polynomials $p_1(x),\ldots,p_k(x)$ and $q(x)$ in $\mathbb Q[x]$ such that in order to have $$p_1(x)r_1(x)+\cdots +p_k(x)r_k(x) = q(x)$$ at least one of $r_1(x),\ldots,r_k(x)$ must have at least one irrational coefficient?
That is, $q(x)$ has rational coefficients, and can be expressed as a linear combination of some specific other polynomials that have rational coefficients, but only by multiplying them by some polynomials that do not have rational coefficients.
Is such a situation possible?
This is the same as asking: Is it possible to start with polynomials in $\mathbb Q[x]$, allow them to generate an ideal in $\mathbb R[x]$, and then find within that ideal some polynomial in $\mathbb Q[x]$ which would not have appeared had the ideal been generated over $\mathbb Q[x]$ instead?
Perhaps the first two warm-up questions were misleading; in each of those cases, we rule out the multiplier being irrational altogether. That is, we show that only a rational multiplier is possible. But in the third question, certainly it is possible that the multipliers $r_i(x)$ could live outside of $\mathbb Q[x]$. The question is, can some choice of the $p_i(x)$ and $q(x)$ ever force that to happen?
Note: This has already been pointed out in the comments, so I might as well acknowledge it: For the sake of simplicity, I am intentionally neglecting instances like $p=0$ and $q=1$ (in Question 1) or $p(x)=x^2$ and $q(x)=x$ (in Question 2) in which the condition is impossible and therefore (technically/vacuously) implies anything.
Answer to Question 3: (added later) I think I can answer Question 3. Consider the ideal generated by the $p_i(x)$ in $\mathbb R[x]$. Since $\mathbb R[x]$ is a PID, this ideal is principal. Now I think it suffices (by Question 2, actually) to show that the generator of this ideal can be taken in $\mathbb Q[x]$. But the most natural generator is the $\rm gcd$ of all the polynomials $p_i(x)$, and this is in fact in $\mathbb Q[x]$. So every polynomial in this ideal is a multiple of this generator, which is rational, and hence by Question 2 every rational polynomial in the ideal is a rational multiple of the generator, and hence has to be in the ideal that the $p_i(x)$ would generate over $\mathbb Q[x]$.
Since I have (I think) answered Question 3, let me move on to...
The Real Question: If in Question 3 we replace $\mathbb Q[x]$ and $\mathbb R[x]$ by $\mathbb Q[x_1,\ldots,x_k]$ and $\mathbb R[x_1,\ldots,x_k]$, does that change the answer?
Some comments give hints to solutions using commutative algebra. Here's a solution using only linear algebra:
consider a complement $\mathbb Q$-vector space $\mathbb Q\oplus S = \mathbb R$.
Then $\mathbb R[x_1,...,x_n] = \mathbb Q[x_1,...,x_n]\oplus S[x_1,...,x_n]$ as $\mathbb Q$-vector spaces. Now this is a complement of $\mathbb Q[x_1,...,x_n]$ that is stable under each endomorphism $p\mapsto x_ip$. Call this endomorphism $\varphi_i$
If we write $\sum_i r_ip_i = \sum_i p_i(\varphi_1,...,\varphi_n)(r_i)$, then the equality $\sum_i r_ip_i = q$ rewrites as $\sum_i p_i(\varphi_1,...,\varphi_n)(r_i) = q$.
We may now mod out by $S[x_1,...,x_n]$ to get $\sum_i p_i(\varphi_1,...,\varphi_n)(\overline{r_i}) = q$ where $\overline{r_i}\in \mathbb Q[x_1,...,x_n]$, which itself rewrites as $\sum_i \overline{r_i}p_i = q$ in $\mathbb Q[x_1,...,x_n]$ : we are done.