No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example let $$F(x)=\begin{cases}x^2\sin(1/x^{10}),&(x\ne0),
\\0,&(x=0).\end{cases}$$
Then $F$ is differentiable everywhere but $\int_{-1}^1 F'(t)\,dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.
Lemma. If $F:\Bbb R\to\Bbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.
It's enough to prove this:
If $\lambda>0$ then $|F(x)-F(0)|\le\lambda x$ for every $x\ge 0$.
Proof: Let $A$ be the set of $a\ge0$ such that $|F(x)-F(0)|\le \lambda x$ for every $x\in[0,a]$. It's clear that $$A=[0,\alpha] $$for some $\alpha\in[0,\infty]$, and we need only show that $\alpha=\infty$. But if $\alpha<\infty$ then $D_RF(\alpha)=0$ shows that there exists $\delta>0$ with $[\alpha,\alpha+\delta)\subset A$. (Choose $\delta$ so that $|F(\alpha+h)-F(\alpha)|<\frac12\lambda h$ for all $h\in[0,\delta)$.)
And now
Prop. Suppose $F:\Bbb R\to\Bbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+\int_0^x f$$for every $x>0$.
Proof: Define $G(x)=\int_0^x f(t)\,dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.
(Cor. If $f$ is continuous then $F$ is differentiable.)
Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.
If $F:\Bbb R\to\Bbb R$ is convex then $F(x)=F(0)+\int_0^x D_RF$.
Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=\int_0^x F'(t)\,dt$ (where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $\int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
Best Answer
As you say, the $\exp$ part is straightforward, so let's look at the derivative of $$ Q = \int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx. $$
A standard thing to do here is to write this as a sum of two integrals, splitting at some arbitrary (but fixed) point $b$: \begin{align} Q &= \int_{t-\tau(t)}^t \frac{\mu(x)U(x)}{S} \,dx\\ &= \int_{t-\tau(t)}^b \frac{\mu(x)U(x)}{S} \,dx + \int_b^t \frac{\mu(x)U(x)}{S} \,dx\\ &= -\int_b^{t-\tau(t)} \frac{\mu(x)U(x)}{S} \,dx + \int_b^t \frac{\mu(x)U(x)}{S} \,dx\\ \end{align} Now you've got two terms, each of which is amenable to taking derivatives. For the first, you get the derivative $$ -\frac{\mu(t-v(t))U(t-v(t))}{S}\left(1 - v'(t) \right). $$ where that final factor comes from applying the chain rule. For the second, you get just $$ \frac{\mu(t)U(t)}{S}. $$
And that's the end of the story.
To answer the question you asked, though, suppose we define $$ H(x) = \frac{\mu(x)U(x)}{S} $$ Then you have $$ Q = \int_{t-v(t)}^t H(x) dx $$ to which you can apply the FTC, and the derivative involves a difference of expressions involving $H$, which you can then expand back out --- no need to integrate a product at all.