This is an interesting question! The 'if' part is true and I think the 'only if' part is also (but this would need verification).
Suppose that $E$ is a $SU(n)$-bundle. Given any $SU(n)$-connection form on $E$, its curvature form $\Omega$ is a basic $\mathfrak{su}(n)$-valued 2-form on the principal $SU(n)$-bundle associated to $E$. As such, it can also be understood as a $\mathfrak{su}(n)$-valued closed 2-form on $M$. According to Chern-Weil theory, the first Chern class is given by the cohomology class of $\frac{i}{2\pi} \mathrm{tr}\,\Omega$, which vanishes since the matrices in $\mathfrak{su}(n)$ are traceless.
Suppose that $E$ is $U(n)$-bundle with vanishing first Chern class. The structure group of $E$ can be reduced to $SU(n)$ if and only if there exists a $SU(n)$-principal subbundle in the $U(n)$-principal bundle $P$ associated to $E$, which is equivalent to the existence of an 'equivariant' function $f : P \to U(n)/SU(n)$ (with the canonical left $U(n)$-action on the quotient). In turn, this is equivalent to the existence of a section of the associated fiber bundle $P(U(n)/SU(n))$. Since $SU(n)$ is a normal subgroup of $U(n)$, we deduce that $Q = P(U(n)/SU(n))$ is a principal bundle, so it admits a section if and only if it is trivial. In fact, $U(n)/SU(n)$ being isomorphic to $U(1)$, it is a $U(1)$-principal bundle. Letting $U(1)$ act on $\mathbb{C}$ in the standard way, we can form the complex line bundle $F = Q(\mathbb{C})$. Thus, the reduction of the structure group of $E$ from $U(n)$ to $SU(n)$ is equivalent to the line bundle $F$ being trivial, that is, as you pointed out, equivalent to $F$ having a vanishing first Chern class.
I think, but I am not quite sure, that looking at how a connection form on $P$ induces a connection form on $Q$, we can infer that the vanishing of the first Chern class of $E$ implies the vanishing of the first Chern class of $F$. If true, the reduction to $SU(n)$ would be guaranteed.
Note that $c_1$ in fact lies in $H^{1,1}(X,\mathbb{Z}):=H^2(X,\mathbb{Z})\cap H^{1,1}(X)$. Then, this is precisely the content of Lefschetz' (1,1)-theorem. See the Wikipedia article on it: https://en.wikipedia.org/wiki/Lefschetz_theorem_on_(1,1)-classes
A great reference for this subject is Huybrechts' book "Complex geometry".
Best Answer
One reference is Principles of Algebraic Geometry by Griffiths and Harris, namely part $2$ of the Proposition on page $141$. To see why that statement is equivalent to the one in your post, you will need to understand the divisor-line bundle correspondence which is explained in the preceding pages. In particular, see the first paragraph on page $136$.