We prove that, for all positive integers $n_1, \ldots, n_m$, $n_{m+1}=n_1$, and non-negative integers $j$ and $r$ with $j\leqslant m$, the following two expressions \begin{align*} &\frac{1}{[n_1+n_m+1]}{n_1+n_{m}\brack n_1}^{-1}\sum_{k=0}^{n_1} q^{j(k^2+k)-(2r+1)k}[2k+1]^{2r+1}\prod_{i=1}^m {n_i+n_{i+1}+1\brack n_i-k},\\[5pt] &\frac{1}{[n_1+n_m+1]}{n_1+n_{m}\brack n_1}^{-1}\sum_{k=0}^{n_1}(-1)^k q^{{k\choose 2}+j(k^2+k)-2rk}[2k+1]^{2r+1}\prod_{i=1}^m {n_i+n_{i+1}+1\brack n_i-k} \end{align*} are Laurent polynomials in $q$ with integer coefficients, where $[n]=1+q+\cdots+q^{n-1}$ and ${n\brack k}=\prod_{i=1}^k(1-q^{n-i+1})/(1-q^i)$. This gives a $q$-analogue of some divisibility results of sums and alternating sums involving binomial coefficients and powers of integers obtained by Guo and Zeng. We also confirm some related conjectures of Guo and Zeng by establishing their $q$-analogues. Several conjectural congruences for sums involving products of $q$-ballot numbers $\left({2n\brack n-k}-{2n\brack n-k-1}\right)$ are proposed in the last section of this paper.