Under your conditions, $P(|r|\leq\frac12\big(1+\frac1{n^c}\big)b^2T)$	will usually go to $0$ as $n\to\infty$. So, the lower bound $1-\frac1{n^2}$ on this probability will be impossible. 

Indeed, for simplicity let $T=1$. Note that $c_b^2:=Var(v_i)=(b^2-1)/12$. Let $u_{ij}:=t_i t_j$, where $t_1,\dots,t_n$ are iid Rademacher random variables (independent of the $v_i$'s), with $P(t_i=\pm1)=1/2$. Let $n\to\infty$. Then 
\begin{equation}
	\frac rn=\Big(\frac1{\sqrt n}\sum_{i=1}^n t_i v_i\Big)^2\to c_b^2 Z^2
\end{equation}
in distribution, 
by the central limit theorem, where $Z\sim N(0,1)$. 
So, for any real $\delta>0$ and all $n>b^2/\delta$,
\begin{equation}
	P(|r|\leq\tfrac12\big(1+\tfrac1{n^c}\big)b^2T)
	\le P(\tfrac{|r|}n\le\tfrac{b^2}n)
	\le P(\tfrac{|r|}n\le\delta)\to P(c_b^2 Z^2\le\delta),
\end{equation}
and the latter probability goes to $0$ 
as $\delta\downarrow0$. 

------------

**Addition in response to the OP's clarifications of the question:**

For simplicity, let $T=1$. 
Suppose first that the $u_{ij}$'s are nonrandom numbers. 
Then 
\begin{equation}
	Er^2=\sum_{i,j,k,\ell} u_{ij}u_{k,\ell}Ev_iv_jv_kv_\ell=\sum_i u_{ii}^2\mu_4
	+2\sum_{i\ne \ell} u_{ii}(u_{i\ell}+u_{\ell i})\mu_3\mu_1
\end{equation}
\begin{equation}
	+\sum_{i\ne k} [u_{ii}u_{kk}+u_{ik}(u_{ik}+u_{ki})]\mu_2^2
	+\sum_{i\ne k\ne \ell}[2u_{ii}u_{k\ell}+(u_{ik}+u_{ki})(u_{i\ell}+u_{\ell i})]\mu_2\mu_1^2
	+\sum_{i\ne j\ne k\ne \ell}u_{ij}u_{k\ell}\mu_1^4, 
\end{equation}
where $\mu_p:=Ev_1^p$, $\sum_{i\ne k\ne \ell}$ denotes the sum over all triples $(i,k,\ell)$ of pairwise distinct $i,k,\ell$, and $\sum_{i\ne j\ne k\ne \ell}$ denotes the sum over all quadruples $(i,j,k,\ell)$ of pairwise distinct $i,j,k,\ell$. 
Note also that 
\begin{equation}
	\sum_{i\ne k\ne \ell}u_{ii}u_{k\ell}
	=\sum_{k\ne \ell}u_{k\ell}\sum_i u_{ii}-\sum_{k\ne \ell}u_{kk}(u_{k\ell}+u_{\ell k}).
\end{equation}

Suppose now further that the $u_{ij}$'s are in $[-1,1]\setminus(-1+1/n,1-1/n)$ for all $i,j$ and such that 
\begin{equation}
	\sum_i u_{ii}=0,\quad u_{ij}+u_{ji}=0 \text{ if $j\ne i$, }\tag{*}
\end{equation}
so that
\begin{equation}
	\sum_{i\ne j} u_{ij}=0\quad\text{and hence}\quad \sum_{i,j} u_{ij}=0.
\end{equation}
E.g., if $n=2m$ is even, we can take $u_{ii}=1$ for $i\le m$, $u_{ii}=-1$ for $i>m$, $u_{ij}=1$ if $i<j$, $u_{ij}=-1$ if $i>j$. Let us consider this case in detail. We have 
\begin{equation}
	Er=\sum_{i,j} u_{ij}Ev_iv_j=\sum_i u_{ii}\mu_2+\sum_{i\ne j} u_{ij}\mu_1^2=0+0=0. 
\end{equation} 
So, 
\begin{equation}
	Var(r)=Er^2=n\mu_4+\sum_{i\ne k} u_{ii}u_{kk}\mu_2^2
	+\sum_{i\ne j\ne k\ne \ell}u_{ij}u_{k\ell}\mu_1^4. 
\end{equation}
Next, 
$\sum_{i\ne k} u_{ii}u_{kk}=\sum_{i,k} u_{ii}u_{kk}-\sum_i u_{ii}^2
=(\sum_i u_{ii})^2-\sum_i u_{ii}^2=0-n=-n$. 
So,
\begin{equation}
	[Var(r)=]Er^2=n(\mu_4-\mu_2^2)
	+\sum_{i\ne j\ne k\ne \ell}u_{ij}u_{k\ell}\mu_1^4. 
\end{equation}
Repeating this reasoning with $1$ in place of $v_i$, we get 
\begin{equation}
	0=\Big(\sum_{i,j} u_{ij}\Big)^2=n(1^4-(1^2)^2)
	+\sum_{i\ne j\ne k\ne \ell}u_{ij}u_{k\ell}1^4, 
\end{equation}
whence  
$\sum_{i\ne j\ne k\ne \ell}u_{ij}u_{k\ell}=0$ and 
\begin{equation}
	Var(r)=Er^2=n(\mu_4-\mu_2^2)=n\,Var(v_1^2)=\tfrac{n}{180} (2 b+1) (8 b+11) \left(b^2-1\right).  
\end{equation}
This is smaller than $n^4b^4T^{2}=n^4b^4$ by a factor of $\asymp n^3$, not just $\asymp n^2$. 
The case of $n$ odd should be very similar. 

Now of course you can take any random $u_{ij}$ (independent of the $v_i$'s) such that $(*)$ holds almost surely or with high enough probability. Is this "packing"/"forcing" condition broad enough?