Here is a tentative proof:
We first notice that by Kummer theorem $\binom{2n}{2m} \equiv \binom{n}{m} \bmod 2$, so that it suffices to show that $$ S_q(n) \equiv T_q(n) \pmod 2$$ where
\begin{align*}
S_q(n):=&\sum_k \binom{k}{n-k}\binom{n+3q}{2k+2q+1}\\
T_q(n):=&\binom{n+3q}{2q-1}.
\end{align*}
From here, the symbol $\sim$ means has the same parity as.
We have
$\binom{a+1}{b+1}\sim \binom{a}{b+1}+\binom{a}{b}$ but also $\binom{a+2}{b+2}\sim \binom{a+2}{b}+\binom{a}{b}$.
We have
\begin{align*} T_q(n+2)&\sim \binom{n+3q+2}{2q-1} \sim \binom{n+3q}{2q-1}+\binom{n+3q}{2q-3}\\
&\sim T_q(n) +\binom{n+3+3(q-1)}{2(q-1)-1} \\
&\sim T_q(n) +T_{q-1}(n+3) \\
T_q(n+3)&\sim T_{q+1}(n+2)+ T_{q+1}(n)
\end{align*}
On the other hand, we have
\begin{align*}S_q(n+3)& \sim \sum_k \binom{k}{n+3-k}\binom{n+3q+3}{2k+2q+1}\\
&\sim \sum_k \binom{k}{n+2-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&+ \sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3q+1}{2(k-1)+2q+3}\\
&+ \sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)+1}\\
&+ \sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
& \sim \sum_k \binom{k-1}{n+2-(k-1)}\binom{n+2+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k\binom{k-1}{n+2-(k-1)}\binom{n+2+3(q+1)-2}{2(k-1)+2(q+1)-1}\\
&+\sum_k \binom{k-1}{n+1-(k-1)}\binom{n+2+3q+1}{2k+2q+1}\\
&\sim S_{q+1}(n+2) +\sum_k\binom{k}{n+2-k}\binom{n+3(q+1)}{2k+2(q+1)-1}\\
&+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2) \\&+\sum_k\binom{k}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+\sum_k\binom{k-1}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k\binom{k-1}{n-(k-1)}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}\\
&+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n) \\&+\sum_k\binom{k-1}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n)+2\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n) \\&+\sum_k\binom{k-1}{n+2-k}\binom{n+3(q+1)}{2(k-1)+2(q+1)+1}+\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
&\sim S_{q+1}(n+2)+ S_{q+1}(n)+2\sum_k \binom{k}{n+1-k}\binom{n+3(q+1)}{2k+2(q+1)+1}\\
S_q(n+3)&\sim S_{q+1}(n+2)+ S_{q+1}(n)
\end{align*}
So the parity of both $ S_q(n)$ and $ T_q(n)$ satisfy the same third order linear recurrence on $n$. Then in order to show that $ S_q(n) \sim T_q(n)$, it suffices to show that $ S_q(0) \sim T_q(0)$, $ S_q(1) \sim T_q(1)$ and $ S_q(2) \sim T_q(2)$.
$S_q(0)-T_q(0)$ is clearly an even integer since
$$S_q(0)-T_q(0)= \binom{3q}{q+1}-\binom{3q}{q-1}=2 \binom{3q+1}{q-1}.$$
We have $T_q(1)-S_q(1)= \binom{3q+1}{q+2}-\binom{3q+1}{q-2}$. After some calculation we find $$T_q(1)-S_q(1)= 2\frac{5q+7}{3q+3}\binom{3q+3}{q-1}$$ which is an even integer since \begin{align*}\binom{3q+3}{q-1}(5q+7) &\equiv \binom{3q+3}{q-1}(2q+4)\pmod {3q+3}\\
&\equiv-\binom{3q+3}{q-1}(q-1)\pmod {3q+3}\\
&\equiv-\binom{3q+2}{q-2}(3q+3)\pmod {3q+3}\\
&\equiv 0\pmod {3q+3}\end{align*}
Also, after some calculation, it can be shown that
\begin{align*}
T_q(2)-S_q(2) &= \binom{3q+2}{q+3}-\binom{3q+2}{q-1}-\binom{3q+2}{q-3}\\
&= 2 \binom{3q+5}{q-1} \frac{54+301q+258q^2+59q^3}{(3q+5)(3q+4)(3q+3)}.
\end{align*}
Then, to complete the proof, we need to show that
$$ \binom{3q+5}{q-1} (54+301q+258q^2+59q^3)\equiv 0 \pmod {(3q+5)(3q+4)(3q+3)}.$$
We have $$ (54+301q+258q^2+59q^3) \equiv (q-1)(66+47q+5q^2)\pmod {(3q+5)(3q+4)(3q+3)}.$$
Then $$ \binom{3q+5}{q-1} (54+301q+258q^2+59q^3) \equiv (3q+5)\binom{3q+4}{q-2}(66+47q+5q^2)\pmod {(3q+5)(3q+4)(3q+3)}.$$
Then it suffices to show that
$$ \binom{3q+4}{q-2}(66+47q+5q^2)\equiv 0 \pmod {(3q+4)(3q+3)}.$$
But $3q+4$ and $3q+3$ are coprime, so we need to show that
$$ \binom{3q+4}{q-2}(66+47q+5q^2)\equiv 0 \pmod {3q+4}$$
and
$$ \binom{3q+4}{q-2}(66+47q+5q^2)\equiv 0 \pmod {3q+3}.$$
That is
$$ \binom{3q+4}{q-2}(14-q^2)\equiv 0 \pmod {3q+4} \tag1$$
and
$$ \binom{3q+4}{q-2}(24-q-q^2)\equiv 0 \pmod {3q+3}.\tag2$$
(1) is equivalent to
\begin{align*} \frac{3q+4}{q-2}\binom{3q+3}{q-3}(14-q^2)&\equiv 0 \pmod {3q+4} \\
\binom{3q+3}{q-3}(14-q^2)&\equiv 0 \pmod {q-2}\\
10\binom{3q+3}{q-3}&\equiv 0 \pmod {q-2}\\
10\frac{q-2}{3q+4}\binom{3q+4}{q-2}&\equiv 0 \pmod {q-2}\\
10\binom{3q+4}{q-2}&\equiv 0 \pmod {3q+4}.\\
\end{align*}
But the last line holds true indeed, because we know from here that
\begin{align*} \binom{3q+4}{q-2}&\equiv 0 \pmod {\frac{3q+4}{\gcd(3q+4,q-2)}}\\
&\equiv 0 \pmod {\frac{3q+4}{\gcd(10,q-2)}}\end{align*}
There remains to show the validity of (2). We have
$$ \binom{3q+4}{q-2}(24-q-q^2)= (3q+3)\binom{3q+4}{q-4} \frac{3q+4}{(q-2)(q-3)}(24-q-q^2)$$
so that we need to show that
$$ \binom{3q+2}{q-4}(3q+4)(24-q-q^2)\equiv 0\pmod {(q-2)(q-3)}. $$
That is
$$ \binom{3q+2}{q-4}(3q+4)(24-q-q^2)\equiv 0\pmod {q-2} $$
and
$$ \binom{3q+2}{q-4}(3q+4)(24-q-q^2)\equiv 0\pmod {q-3}. $$
That is
$$ 180\binom{3q+2}{q-4}\equiv 0\pmod {q-2} $$
and
$$ 156 \binom{3q+2}{q-4}\equiv 0\pmod {q-3}. $$
That is
$$ 180\frac{(q-2)(q-3)}{(3q+4)(q-2)}\binom{3q+4}{q-2}\equiv 0\pmod {q-2} $$
and
$$ 156 \frac{q-3}{3q+3}\binom{3q+3}{q-3}\equiv 0\pmod {q-3}. $$
That is
$$ 180(q-3)\binom{3q+4}{q-2}\equiv 0\pmod {(3q+4)(3q+3)} $$
and
$$ 156\binom{3q+3}{q-3}\equiv 0\pmod {3q+3}. $$
That is
$$ 180(q-3)\binom{3q+4}{q-2}\equiv 0\pmod {3q+4} \tag3$$
and
$$ 180(q-3)\binom{3q+4}{q-2}\equiv 0\pmod {3q+3} \tag4$$
and
$$ 156\binom{3q+3}{q-3}\equiv 0\pmod {3q+3}. \tag5$$
For (5) it holds true because we have $156=13\cdot 12$ a multiple of $12$ and from here, we have
\begin{align*}\binom{3q+3}{q-3}&\equiv 0\pmod {\frac{3q+3}{\gcd(3q+3,q-3)}} \\
&\equiv 0\pmod {\frac{3q+3}{\gcd(12,q-3)}} \end{align*}
For (3) it holds true because we have $180=18\cdot 10$ a multiple of $10$ and from here, we have
\begin{align*}\binom{3q+4}{q-2}&\equiv 0\pmod {\frac{3q+4}{\gcd(3q+4,q-2)}} \\
&\equiv 0\pmod {\frac{3q+2}{\gcd(10,q-2)}} \end{align*}
We have seen that there exist an integer $K$, such that $\binom{3q+3}{q-3}=K\frac{3q+3}{\gcd(12,q-3)}$. Now, with the same argument from here, we have
\begin{align*}\binom{3q+3}{q-2}&\equiv 0\pmod {\frac{3q+3}{\gcd(3q+3,q-2)}} \\
&\equiv 0\pmod {\frac{3q+3}{\gcd(9,q-2)}} \end{align*}
so there exists an integer $L$ such that $\binom{3q+3}{q-2}=L\frac{3q+3}{\gcd(9,q-2)}$.
Then
\begin{align*} 180(q-3)\binom{3q+4}{q-2}&=180(q-3)\left(\binom{3q+3}{q-2}+\binom{3q+3}{q-3}\right)\\
&= (3q+3)\left( \frac{180L(q-3)}{\gcd(9,q-2)}+\frac{180K(q-3)}{\gcd(12,q-3)}\right). \end{align*}
The second factor on the right hand side is clearly an integer and this establishes the validity of (4). The proof is finished here.