Riemann Integrability and Jordan Measure
Take the function $f :[0,1] \to \mathbb{R}$ where
$f(x)=\begin{cases}\ 1 & x \in \mathbb{Q}\cap [0,1]\\ 0 & x \in \mathbb{Q}^c \cap [0,1] \end{cases}$
The graph of this function has Jordan content zero which implies that it has Jordan measure zero.
But $f$ is not Riemman integrable.
Initially I was going to answer your question in the affirmative but I really could not find any flaw in the answers already given so I stopped for a while to figure out why I wanted to answer in the affirmative. This is what I found out.
Let us start with the following theorem
Theorem: Let $A$ be a non empty and bounded subset of $\mathbb{R} \times \mathbb{R} $. Then $A$ is Jordan measurable if and only if the boundary of $A$ has Jordan measure $0$.
And Riemann integration is intimately linked with Jordan measure via the following theorem
Theorem: Let $f:[a, b] \to\mathbb{R} $ be a non-negative function and let $$A=\{(x, y) \mid x\in[a, b], 0\leq y\leq f(x) \} $$ be its subgraph (area below the graph). Then $A$ is Jordan measurable if and only $f$ is Riemann integrable on $[a, b] $ and further Jordan measure of $A$ is equal to the Riemann integral $\int_{a}^{b} f(x) \, dx$.
I was well aware of these two theorems stated above and then from these one can conclude that
The Riemann integral of $f$ over $[a, b] $ exists if and only if the boundary of subgraph of $f$ is of Jordan measure $0$.
The boundary of subgraph of $f$ has four parts: two ordinates at $a$ and $b$ of lengths $f(a) $ and $f(b) $, part of $x$ axis from $a$ to $b$ and finally the graph of $f$. The first three parts have Jordan measure $0$ and hence it follows that the Riemann integral of $f$ over $[a, b] $ exists if and only if the graph of $f$ is of Jordan measure $0$.
This is wrong!! The subtle flaw in the argument of above paragraph is about the fourth component of the boundary of the region below graph of $f$. It may not just be the graph of $f$ but rather include more points if the function is too discontinuous. Thus if $f$ is the characteristic function of rationals on $[0,1] $ the whole subgraph of $f$ is the boundary and it has outer Jordan measure $1$. The graph of $f$ here is still of Jordan measure $0$.
HINT:
The graph of a function with a finite image has Jordan measure $0$.
$\bf{Added:}$ After we have seen that the answer to the OP question is negative ( since every function with say an image $\subset \mathbb{R}$ of Jordan measure $0$ has also the graph of Jordan measure $0$), there is still a result to be answered in the affirmative. Assume for simplicity that $f$ is positive. Then $f$ is Riemann intégrable if and only if the subgraph $\{ y \le f(x) \}$ is Jordan measurable, and the integral of $f$ is the measure of the subgraph. The proof is not difficult.