Define the sequence $a_n(x):=\sum_{j=0}^n(-2)^{n-j}\binom{n}j\binom{n+j}jx^j$ so that
$r(n)=\int_0^2\frac{a_n(x)-a_n(1)}{x-1}dx$.

We need the following fact which follows from the Vandermonde-Chu identity: for $n\geq1$,
\begin{align}\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1}
&=\sum_{j=0}^n\binom{n}{n-j}\binom{-n-1}j\frac1{j+1} \\
&=\frac1{n+1}\sum_{j=0}^n\binom{n+1}{n-j}\binom{-n-1}j=0. \tag1
\end{align}
At present, the implication of (1) is that
$$\int_0^2a_n(x)dx=2(-2)^n\sum_{j=0}^n(-1)^j\binom{n}j\binom{n+j}j\frac1{j+1}=0. \tag2$$
Next, we apply Zeilberger's algorithm to generate the two recursive relations
\begin{align}
(n+2)a_{n+2}(x)-2(x-1)(2n+3)a_{n+1}(x)+4(n+1)a_n(x)&=0, \\
(n+2)a_{n+2}(1)\qquad \qquad \qquad \qquad \qquad \qquad+4(n+1)a_n(1)&=0.
\end{align}
Subtract the $2^{nd}$ equation from the $1^{st}$, divide through by $x-1$ and integrate $\int_0^2(\cdot)dx$. So,
$$(n+2)r(n+2)+4(n+1)r(n)=0\qquad \implies \qquad r(n+2)=-\frac{4(n+1)}{n+2}\,r(n);$$
where (2) has been utilized effectively. Initial conditions are $r(0)=0$ and $r(1)=4$. The case $n$ even is transparent. The case $n$ odd also follows from induction on $n$ and the fact that if we write the RHS of your claim for $r(n)$ as
$$t(n):=\frac{(-1)^{\lfloor n/2\rfloor}4^n}{n\binom{n-1}{\lfloor n/2\rfloor}}\chi_{odd}(n)$$
then it is easy to check that $\frac{t(n+2)}{t(n)}=-\frac{4(n+1)}{2n+3}$ when $n$ is odd. The proof is now complete.

**Note.** Here $\chi_{odd}$ is understood as $\chi_{odd}(odd)=1$ and $\chi_{odd}(even)=0$.