Let $M$ be an $n$-manifold endowed with a nonvanishing $n$-form $\Omega$, let $\mathrm{Diff}(M,\Omega)$ denote the group of $\Omega$-preserving diffeomorphisms of $M$, and, for $x\in M$, let $\mathrm{Diff}(M,\Omega,x)$ denote the subgroup that fixes $x$.  

Your question, then, easily reduces to "Is the homomorphism $D:\mathrm{Diff}(M,\Omega,x)\to \mathrm{SL}(T_xM)$ defined by $D(f) = f'(x):T_xM\to T_xM$ surjective?"

The answer is 'yes'.  The reason is that the image of $D$ has to be a Lie subgroup of $\mathrm{SL}(T_xM)$, and it is very easy to see that it contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$.  Consequently, it must be all of $\mathrm{SL}(T_xM)$.

**Added at the request of the OP**: Showing that the image of the homomorphism $D$ contains a neighborhood of the identity in $\mathrm{SL}(T_xM)$ can be done in a number of ways, but here is one that is relatively 'low-tech': Note that the result is obvious when $n=1$, so assume that $n>1$ from now on.  Then, one can choose coordinates $u=(u^i)$ centered on $x\in M$ and defined on a neighborhood $U$ of $x$ in which
$$
\Omega = \mathrm{d}u^1\wedge\mathrm{d}u^2\wedge\cdots\wedge\mathrm{d}u^n = \mathrm{d}u
$$
There is an $r>0$, so that $u(U)\subset\mathbb{R}^n$ contains the open ball $B_r(0)$ of radius $r$ about $0\in\mathbb{R}^n$.  Let $G_0 = \mathrm{Diff}_0\bigl(B_r(0),\mathrm{d}u,0\bigr)$ denote the group of diffeomorphisms $\phi:B_r(0)\to B_r(0)$ that satisfy $\phi(0)=0$, $\phi^*(\mathrm{d}u) = \mathrm{d}u$, and for which there exists a compact set $K_\phi\subset B_r(0)$ such that $\phi$ is the identity outside the compact set $K_\phi$.  Clearly, it will be enough to show that the homomorphism $D:G_0\to \mathrm{SL}\bigl(T_0B_r(0)\bigr) = \mathrm{SL}(\mathbb{R}^n)$ is surjective.

To do this, consider the subset $C\subset \mathrm{Hom}_0(\mathbb{R}^n,\mathbb{R}^n)={\frak{sl}}(n,\mathbb{R})$ consisting of the operators with trace zero whose eigenvalues are purely imaginary and either all nonzero (if $n$ is even) or one zero with multiplicity 1 (if $n$ is odd). Note that $C$ is an open cone in ${\frak{sl}}(n,\mathbb{R})$.  It is the interior of the set of operators $L:\mathbb{R}^n\to\mathbb{R}^n$ for which the $1$-parameter subgroup $\mathrm{e}^{tL}$ has compact closure in $\mathrm{SL}(n,\mathbb{R})$.  In particular, for such an operator, there is at least one positive definite inner product $\langle,\rangle$ on $\mathbb{R}^n$, such that the flow of $\mathrm{e}^{tL}$ preserves the inner product $\langle,\rangle$ and hence the volume form $\mathrm{d}u$.  Now, let $\epsilon>0$ be so small that the set of vectors $y\in\mathbb{R}^n$ that satisfy $\langle y,y\rangle \le \epsilon$ is a compact subset of $B_r(0)$.  Let $h:\mathbb{R}\to\mathbb{R}$ be a smooth function that is identically $1$ when $0\le t \le \epsilon/2$ and vanishes identically for $t\ge\epsilon$.  Now consider the $1$-parameter family of smooth maps
$$
f_t(y) = \mathrm{e}^{t\ h(\langle y,y\rangle) L}y.
$$
This family preserves the level sets of $Q(y) = \langle y,y\rangle$, which are spheres, and rigidly rotates each one, so it preserves the volume form $\mathrm{d}u$.  It clearly lies in $G_0$.  When $Q(y)\ge\epsilon$, $f_t(y) = y$, and, when $Q(y)\le \epsilon/2$, we have $f_t(y) = \mathrm{e}^{tL}y$.  Thus, $f_t$ lies in $G_0$.  Thus, since
$$
D(f_t) = \mathrm{e}^{tL},
$$
it follows that the image of $D$ contains the subset $\mathrm{e}^{C}\subset \mathrm{SL}(n,\mathbb{R})$, which obviously has non-empty interior.  This subset is also closed under inverse, since $C = -C$, and from this, it follows that $\mathrm{e}^{C}$ generates $\mathrm{SL}(n,\mathbb{R})$.  In particular, it follows that the image of $D$ is all of $\mathrm{SL}(n,\mathbb{R})$, which is what needed to be proved.