Decompose $XX^T = O^T \Lambda O$ with $O$ an $M\times M$ orthogonal matrix and $\Lambda={\rm diag}\,(\lambda_1,\lambda_2,\ldots \lambda_M)$ the diagonal matrix of eigenvalues. Define $w=|v|^{-1} Ov$, then
$$v^T XX^T v =|v|^2 \sum_{m=1}^M \lambda_m w_m^2.$$
The matrix $XX^T$ has a <A HREF="https://en.wikipedia.org/wiki/Wishart_distribution">Wishart distribution,</A> with independent $O$ and $\Lambda$. It follows that the $w_m$'s are independent Gaussians with mean zero and variance $1/M$. The probability distribution of the $\lambda_m$'s is
$$P(\lambda_1,\lambda_2,\ldots\lambda_M)\propto \prod_{m=1}^M e^{-\lambda_m/2}\lambda_m^{(N-M-1)/2}\prod_{i<j}|\lambda_i-\lambda_j|,$$
with $E[\sum_{m}\lambda_m]=NM$.

This gives
$$E\left[ {{v^T}X{X^T}{v}} \right]=|v|^2 N.$$
The expectation $E\left[\exp(- {{v^T}X{X^T}{v}}) \right]$ can be evaluated by integration for small $M$, 
$$E\left[\exp(- {{v^T}X{X^T}{v}}) \right]=\int_0^\infty d\lambda_1\cdots\int_0^\infty d\lambda_M \,P(\lambda_1,\ldots\lambda_M)\prod_{m=1}^M(1+2M|v|^2\lambda_m)^{-1/2},$$
for large $M$ it tends to $e^{-|v|^2 N}$.