No: given $\eta > \epsilon > 0$, there are $n \times n$ symmetric matrices $A_n$
with spectral radius $> \eta$, such that $Pr\left[|v^TA_nv|/\|v\|^2 > \epsilon\right] < e^{-cn}$ for some $c > 0$.

I assume a standard Gaussian distribution, with mean $0$ and covariance matrix $I$.
Consider an $n \times n$ diagonal matrix $A$ with one diagonal element $\alpha$ and
the other diagonal elements $-\epsilon$, where $\alpha > \eta > \epsilon > 0$. Then
$v^T A v = \alpha v_1^2 - \epsilon \sum_{j=2}^{n} v_j^2$, and it is impossible to have
$v^T A v < -\eta \|v\|^2$, while
$$\eqalign{Pr\left[v^T A v > \eta \|v\|^2 \right] &=
Pr\left[ (\alpha - \eta) v_1^2 - \sum_{j=2}^n (\epsilon + \eta) v_j^2 > 0 \right]\cr
&\le Pr \left[ (\alpha - \eta) v_1^2 > k (n-1)\right] +
Pr\left[ S_n < k (n-1)\right]\cr}$$
for any $0 < k < \epsilon + \eta$, where $S_n = \sum_{j=2}^n (\epsilon + \eta) v_j^2$.

Now $Pr[(\alpha - \eta) v_1^2 > k (n-1)]$ goes to 0 exponentially as $n \to \infty$.
On the other hand,
$S_n$ has mean $(n-1)(\epsilon+\eta)$, and for any
$k < \epsilon + \eta$, $Pr[S_n < k (n-1)]$ goes to $0$ exponentially by the theory of large deviations.