You can calculate the variance for the percentage change as follows: Your problem is to calculate $\mathrm{Var}(100\cdot (Y-X)/X)$. This can be written as: $100^{2}\mathrm{Var}(Y/X)$. Now we can use the delta method to calculate the approximative variance. The delta method states that an approximation of the variance of a function $g(t)$ is given by:
$$
\mathrm{Var}(g(t))\approx \sum_{i=1}^{k}g'_{i}(\theta)^{2}\mathrm{Var}(t_{i})+2\sum_{i>j}g'_{i}(\theta)g'_{j}(\theta)\mathrm{Cov}(t_{i},t_{j})
$$
Your function $g(t)$ is: $g(\mu_{X},\mu_{Y})=\mu_{Y}/\mu_{X}$. For the variance, we need the partial derivatives of $g(\mu_{X},\mu_{Y})$:
$$
\begin{align}
\frac{\partial}{\partial \mu_{X}}g(\mu_{X},\mu_{Y}) & = -\frac{\mu_{Y}}{\mu_{X}^{2}} \\
\frac{\partial}{\partial \mu_{Y}}g(\mu_{X},\mu_{Y}) &= \frac{1}{\mu_{X}} \\
\end{align}
$$
Using the function for the variance above, we get:
$$
\mathrm{Var}(Y/X)\approx\left(\frac{\mu_{Y}^{2}}{\mu_{X}^{4}}\right)\mathrm{Var}(X) + \left(\frac{1}{\mu_{X}^{2}}\right)\mathrm{Var}(Y)-2\cdot \left(\frac{\mu_{Y}}{\mu_{X}^{3}}\right)\mathrm{Cov}(X,Y)
$$
So the estimated standard error of the percentage change would be:
$$
\mathrm{SE}(100\cdot (Y-X)/X)\approx100\cdot \sqrt{\frac{\mathrm{Var}(Y)\mu_{X}^{2}-2\cdot\mathrm{Cov}(X,Y)\mu_{X}\mu_{Y}+\mathrm{Var}(X)\mu_{Y}^{2}}{\mu_{X}^4}}
$$
If $X$ and $Y$ are independent (i.e. $\mathrm{Cov}(X,Y)=0$), the formula simplifies to:
$$
\mathrm{SE}(100\cdot (Y-X)/X)\approx100\cdot \sqrt{\frac{\mathrm{Var}(Y)\mu_{X}^{2}+\mathrm{Var}(X)\mu_{Y}^{2}}{\mu_{X}^4}}
$$
