The expectation can be computed in closed form, and I think that without further assumptions on entries of the matrix $Y$, the Jensen bound is sharp according to following calculation:
$\begin{align}\mathbb{E}\left[\boldsymbol{X^{t}YX}\right] & =\mathbb{E}\left[\sum_{\substack{1\le i,j\le n}
}X_{i}X_{j}Y_{ij}\right]\\
& =\sum_{m=-n}^{n}\mathbb{E}\left[\sum_{1\le i,j\le n}X_{i}X_{j}Y_{ij}\left|\sum_{i=1}^{n}X_{i}=m\right.\right]\mathbb{P}\left(\sum_{i=1}^{n}X_{i}=m\right)\\
& =\sum_{m=-n}^{n}\sum_{1\le i,j\le n}Y_{ij}\mathbb{E}\left[X_{i}X_{j}\left|\sum_{i=1}^{n}X_{i}=m\right.\right]\mathbb{P}\left(\sum_{i=1}^{n}X_{i}=m\right)\\
& =\sum_{m=-n}^{n}\sum_{1\le i,j\le n}Y_{ij}\left\{ x_{i}x_{j}\mathbb{P}\left(X_{i}=x_{i},X_{j}=x_{j}\left|\sum_{i=1}^{n}X_{i}=m\right.\right)\right\} \mathbb{P}\left(\sum_{i=1}^{n}X_{i}=m\right)\\
& =\sum_{m=-n}^{n}\sum_{1\le i,j\le n}Y_{ij}\left\{ \sum_{x_{i},x_{j}\in\{\pm1\}}x_{i}x_{j}\mathbb{P}\left(\left.\sum_{i=1}^{n}X_{i}=m\right|X_{i}=x_{i},X_{j}=x_{j}\right)\mathbb{P}\left(X_{i}=x_{i},X_{j}=x_{j}\right)\right\} \\
& \text{The probability }\mathbb{P}\left(X_{i}=x_{i},X_{j}=x_{j}\right)=\left(\frac{1}{2}\right)^{x_{i}+x_{j}}\left(\frac{1}{2}\right)^{2-x_{i}-x_{j}}=\frac{1}{4},\forall x_{i},x_{j}\in\{\pm1\}\\
& =\sum_{m=-n}^{n}\sum_{1\le i,j\le n}Y_{ij}\frac{1}{4}\left\{ \sum_{x_{i},x_{j}\in\{\pm1\}}x_{i}x_{j}\mathbb{P}\left(\left.\sum_{i=1}^{n}X_{i}=m\right|X_{i}=x_{i},X_{j}=x_{j}\right)\right\} \\
& \text{The probability }\mathbb{P}\left(\left.\sum_{i=1}^{n}X_{i}=m\right|X_{i}=x_{i},X_{j}=x_{j}\right)=\begin{cases}
m\neq\pm n,\pm(n-2) & \begin{cases}
\frac{\left(\begin{array}{c}
n-2\\
\frac{n-2-m}{2}
\end{array}\right)\left(\frac{1}{2}\right)^{\frac{n-2-m}{2}+m}\left(\frac{1}{2}\right)^{\frac{n-2-m}{2}}}{1/4} & n-m\,even\\
0 & n-m\,odd
\end{cases}\\
m=\pm n & \left(\frac{1}{2}\right)^{n}I_{(x_{i}=x_{j}=1)}\text{ or }\left(\frac{1}{2}\right)^{n}I_{(x_{i}=x_{j}=-1)}\\
m=\pm(n-2) & \left(\frac{1}{2}\right)^{n}I_{(x_{i}=1,x_{j}=-1)}\text{ or }\left(\frac{1}{2}\right)^{n}I_{(x_{i}=-1,x_{j}=1)}
\end{cases}\,\text{treated as a 1-d random walk.}\\
& \text{Following simplification assumes }m\neq\pm n,\pm(n-2)\\
& =\sum_{m=-n}^{n}\sum_{1\le i,j\le n}Y_{ij}\frac{1}{4}\left\{ \sum_{x_{i},x_{j}\in\{\pm1\}}x_{i}x_{j}\left(\frac{1+(-1)^{n-m}}{2}\right)\cdot\frac{\left(\begin{array}{c}
n-2\\
\frac{n-2-m}{2}
\end{array}\right)\left(\frac{1}{2}\right)^{\frac{n-2-m}{2}+m}\left(\frac{1}{2}\right)^{\frac{n-2-m}{2}}}{1/4}\right\} \\
& =\sum_{m=-n}^{n}\sum_{1\le i,j\le n}Y_{ij}\left\{ \sum_{x_{i},x_{j}\in\{\pm1\}}x_{i}x_{j}\left(\frac{1+(-1)^{n-m}}{2}\right)\cdot\left(\begin{array}{c}
n-2\\
\frac{n-2-m}{2}
\end{array}\right)\left(\frac{1}{2}\right)^{\frac{n-2-m}{2}+m}\left(\frac{1}{2}\right)^{\frac{n-2-m}{2}}\right\}
\end{align}
$
If $Y=I_n$ then the lower bound is literally reached. If you want a Hanson-Wright type concentration bound, then it can be improved since Bernoulli random vectors are sub-gaussian:

Hanson-Wright inequality.

Let $X=(X_{1},\cdots X_{n})\in\mathbb{R}^{n}$ be a random vector with
independent components $X_{i}$ such that

(i)$EX_{i}=0$

(ii) $\left\Vert X_{i}\right\Vert _{\psi_{2}}=
sup_{p\geq1}p^{-\frac{1}{2}}\left[E\left|X\right|^{p}\right]^{\frac{1}{p}}\leq K$ i.e. its components have uniform sub-gaussian norm.

Then for arbitrary n\times n constant matrix A and $\forall t\geq 0$
we can assert that

$Pr\left\{ \left|X^{t}AX-EX^{t}AX\right|>t\right\}
\leq2exp\left(-c\cdot min\left(\frac{t^{2}}{K^{4}\left\Vert
A\right\Vert _{HS}^{2}},\frac{t}{K^{2}\left\Vert A\right\Vert
}\right)\right)$ for some constant $c>0$.

where $\left\Vert A\right\Vert =max_{x\neq0}\frac{\left\Vert
Ax\right\Vert _{L^{2}}}{\left\Vert x\right\Vert _{L^{2}}}$ and
$\left\Vert A\right\Vert
_{HS}=\sqrt{\sum_{i,j}\left|a_{ij}\right|^{2}}$.

It is readily verifed that a Bernoulli random vector satisfies (i)(ii)
with $K=2$. Therefore we can assert that

$$Pr\left\{ \left|X^{t}YX-EX^{t}YX\right|>t\right\}
\leq2exp\left(-c\cdot min\left(\frac{t^{2}}{K^{4}\left\Vert
Y\right\Vert _{HS}^{2}},\frac{t}{K^{2}\left\Vert Y\right\Vert
}\right)\right)$$

where $Y=yy^{t}-zz^{t}$ is symmetric as stated in the OP.