I think you may have more restrictive hypotheses in mind, because in general this will be false due to behavior that may be obviously absent in the cases you want to consider. Anyway, here are two counterexamples.
I see $\mathbb S^n$ as the set of $(x_0,\ldots,x_n)\in\mathbb R^{n+1}$ such that $x_0^2+\cdots+x_n^2=1$.
Define $F:(-1,1)\times\mathbb S^n\to\mathbb R^{n+1}$ as
$$ (t,x)\mapsto(tx_0,x_1,\ldots,x_n) $$
(I am squashing the sphere on itself like I eat a burger). Then $F_\varepsilon(\mathbb S^n)$ is basically two copies of a disc glued along their boundaries for $\varepsilon$ very small, but $F_0(\mathbb S^n)$ is just one of those discs, so the volume drops at this time. If you want to make this precise, you can find the surface area of the ellipsoid on the dedicated Wikipedia article, and check that for $\alpha$ such that $\mathcal H^2=\alpha\operatorname{Leb^2}$ over $\mathbb R^2$, the Hausdorff volume of $F_\varepsilon(\mathbb S^2)$ is
$$ 2\alpha\pi\big(1+\varepsilon^2\log(1/|\varepsilon|)+O(\varepsilon^2)\big) $$
for $\varepsilon\neq0$ small in absolute value, and $F_0(\mathbb S^2)=\alpha\pi$.
Let $G:\mathbb S^n\to\mathbb R^{n+1}$ be a smooth embedding such that $G(x)=(0,x_1,\ldots,x_n)$ for all $x$ with $x_0>0$, and $\rho:\mathbb R\to\mathbb R$ a smooth function with $\mathbf1_{\{0\}} \leq\rho \leq \mathbf1_{(-1/2,1/2)}$; these can be constructed by hand. Define
$$ F:(t,x)\mapsto G(x) + t\mathbf1_{x_0>0} \rho(x_1^2+\cdots+x_n^2)\sin(x_1/t^2)e_0 $$
for $t>0$, and $F(t,x)=G(x)$ for $t\leq0$. What is happening is that I am adding ripples on the surface of the sphere, smaller and smaller but also wilder and wilder as $t\to0$.
The continuity of $F$ can be checked sequentially, which becomes easy once one realizes that $\|F_t-F_0\|_\infty\leq |t|$. I claim, and I will not prove, that the volume of $F_\varepsilon(\mathbb S^n)$ is of order $1/\varepsilon$ as $\varepsilon\to0$, $\varepsilon>0$. Clearly the volume cannot be continuous then.
The easiest case $n=1$, which in my opinion already gives a good idea of what is happening, can be treated as follows. For positive $t$, the Hausdorff measure of $F_t(\mathbb S^n)$ is lower bounded by a constant times the length of
$$ \big\{(t\rho(y^2)\sin(y/t^2),y),y\in(-1,1)\big\}. $$
Writing $\gamma_t(y)$ for the first coordinate of this point, this length is
$$\begin{aligned}
\int_{-1}^1\sqrt{1+\gamma_t'(y)^2} \, \mathrm dy
&\geq \int_{-1}^1|\gamma_t'(y)| \,\mathrm dy \\
&\geq \int_{-1}^1\Big(\frac1t\rho(y^2) \left|\cos(y/t^2)\right| - 2t |y\rho'(y^2)\sin(y/t^2)| \Big) \, \mathrm dy \\
&\geq t\inf_{|y|<\delta}\rho(y^2)\cdot\int_{-\delta/t^2}^{\delta/t^2} \left|\cos(y)\right| \, \mathrm dy - 4t\|\rho'\|_\infty \\
&\geq \frac\delta t\inf_{|y|<\delta}\rho(y^2) - 4t\|\rho'\|_\infty \\
\end{aligned}$$
for all $\delta>0$ (the fact that the last integral, say over $[-M,M]$, is at least $M$, is true but not obvious; that said, it is clearly asymptotically $4M/\pi$). Fixing $\delta>0$ small enough that the infimum is positive, we get the expected lower bound of order $1/t$. If we wanted to, we could find an upper bound of the same order along the same lines.
My guess, based on very little (basically just the fact that it rules out both examples above), is that this will be true if the function $(t,x)\to F_t(x)$ is jointly Lipschitz and every $F_t$ is injective (since $\mathbb S^n$ is compact, it imposes that it is actually an embedding, for instance). But I think this is a different question altogether.
