First notice that $10$ is a multiple of $5 = 2 \cdot 2 + 1$, so we are done if $q$ and $10$ are not coprime. If $q$ and $10$ are coprime, since $q$ is prime, by Fermat's theorem $q \mid 10^{q-1} - 1 \Leftrightarrow q \mid 10^{2p} - 1 \Leftrightarrow q \mid 10^p - 1$ or $q \mid 10^p + 1$. If $q \mid 10^p + 1$, we are done: the sum of the digits of $10^p + 1$ is $2$. From now on, suppose that $q \mid 10^p - 1$.

Consider the minimum positive integer $d$ such that $10^d \equiv 1 \pmod{q}$, that is, $d = \text{ord}_q 10$. Then $d \mid p$, that is, the possible values of $d$ are $1$ and $p$. If $d = 1$, $q \mid 10^1 - 1 \Leftrightarrow q = 3$, which is not acceptable. So $d = p$, which means that $1, 10, 10^2, \dots, 10^{p-1}$ are all distinct mod $q$.

Now let $A = \{0\} \cup \{10^k \text{ mod } q : 0 \le k < p\}$ and $B = \{q-1\} \cup \{-1 - 10^k \text{ mod } q : 0 \le k < p\}$. Since $10^k \not\equiv 0 \pmod{q}$, $|A| = |B| = p+1$. This and the obvious fact that $|A \cup B| \le q = 2p+1$ implies $A \cap B \ne \emptyset$.

So either there are integers $r, s$ such that $10^r \equiv -1 - 10^s \pmod{q}$, in which case $10^r + 10^s + 1$, whose digit sum is $3$, is a multiple of $q$, or $10^r \equiv -1 \pmod{q}$, in which $10^r + 1$, whose digit sum is $2$, is a multiple of $q$, or $0 \equiv -1 - 10^s \pmod{q}$, which is analogous to the previous case. So, in all cases, we are done.