Let $\{W_t\}_{t\in[0;T]}$ be a Brownian motion, let $\mu,\sigma\colon [0;T]\times\mathbb R \to \mathbb R$ be continuous, bounded and Lipschitz continuous in the second argument, let $X$ be the unique stochastic process satisfying $$ X_t = \int_0^t \mu^X(s,X_s) \mathrm ds + \int_0^t \sigma^X(s,X_s) \mathrm dW_s. $$ Let $h\colon [0;T]\times \mathbb R \to \mathbb R$ be uniformly continuous and consider $Y_t = h(t,X_t)$. Assume that I already know that $Y$ can be expressed as $$ Y_t = Y_0 + \int_0^t \mu^Y_s \mathrm ds + \int_0^t \sigma^Y_s \mathrm dW_s. $$ with essentially bounded $\mu^Y$ and $\sigma^Y \in \mathcal L^2$. I want to prove that $\sigma^Y$ is essentially bounded or of bounded variation (both would help me) or at least something like $\sigma^Y \in \mathcal L^p$ for $p > 2$.
Example: If $h$ was $C^{1,2}$ with bounded derivatives, I could simply apply Itô's formula and would obtain that $$ \sigma^Y_t = \partial_x h(t,X_t) \sigma^X(t,X_t), $$ which would prove that $\sigma^Y$ is bounded.
But what if $h \not\in C^{1,2}$? Can I obtain a similar (or weaker) result for weaker assumptions?