TL;DR: Yes, OP is correct.

Let's discretize time $t$. In other words, assume that we have $n$ Grassmann-odd variables $\theta^1, \ldots, \theta^n$. In this language OP's differential operator $\partial_t$ is replaced with some Grassmann-even matrix $A^i{}_j$. Define $$\theta^{\prime i}= \sum_{j=1}^nA^i{}_j \theta^j,$$
so that 
$$\frac{\partial}{\partial\theta^j} ~=~\sum_{i=1}^n A^i{}_j\frac{\partial}{\partial\theta^{\prime i}}.$$
Then a straightforward calculation shows that OP's path-integral (1) is replaced with$^1$
$$\begin{align} \int\!d\theta^n \ldots d\theta^1~\theta^{\prime 1}\ldots \theta^{\prime n}~=~&\ldots~=~
\int\! d\theta^n \ldots d\theta^1~\delta(\theta^{\prime 1})\ldots \delta(\theta^{\prime n})\cr
~=~&\ldots~=~\det(A),\end{align} $$
which indeed is a discrete version of OP's claim (2). 

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$^1$Recall that Berezin integration 
$$ \int\!d\theta^n \ldots d\theta^1  ~=~\frac{\partial}{\partial\theta^n}\ldots \frac{\partial}{\partial\theta^1}$$ 
is the same as differentiation!