Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant?

Yes, such an injective morphism has non-zero determinant.

Actually, if $M$ is a finitely generated free module over the commutative ring $R$ and $u:M\to M$ is an endomorphism, one has the precise equivalence: $$u \;\text {is injective}\iff \det(u) \; \text {is not a zero divisor in}\; R.$$ The proof is based on the fact that elements $m_1,m_2, \ldots ,m_n\in M$ form a linearly independent set iff there exists a non-zero $0\neq \lambda\in R$ with $\lambda (m_1\wedge m_2\wedge \ldots\wedge m_n)=0\in \Lambda^nM$.

You can find the details in Bourbaki, Algebra, III, §7, Proposition 3, page 524.

The argument from Bourbaki referenced in Georges Elencwajg's answer is exactly the kind of argument I was hoping would work so let me record it here, with some simplifications.

Proposition: Let $m_1, ... m_n$ be elements of some $R$-module $M$ which are linearly dependent. Then there exists nonzero $r \in R$ such that $r (m_1 \wedge ... \wedge m_n) = 0$ in $\Lambda^n(M)$.

Proof. If $\sum r_i m_i = 0$ is a linear dependence, assume WLOG that $r_1 \neq 0$. Then $r_1 m_1 = - \sum_{i \ge 2} r_i m_i$, so by basic properties of the wedge product we find that $(r_1 m_1) \wedge m_2 \wedge ... \wedge m_n = 0$, so we can take $r = r_1$.

Proposition: If $M$ is free then the converse holds.

Proof. We proceed by induction on $n$. If $n = 1$ this is clear. In general, suppose that there exists a nonzero $r$ such that $$r (m_1 \wedge ... \wedge m_n) = 0.$$

If $r (m_2 \wedge ... \wedge m_n) = 0$, then by the inductive hypothesis $m_2, ..., m_n$ are linearly dependent and we are done, so suppose otherwise. Then $r (m_2 \wedge ... \wedge m_n) \neq 0$. Hence, using the fact that $\Lambda^{n-1}(M)$ is also free (this is crucial!), we conclude that there exists an alternating $n-1$-form $f : M^{n-1} \to R$ such that $$f(r m_2, ..., m_n) = s \neq 0$$ (because of the following fact: if $g$ is a nonzero vector in a free $R$-module $F$, then there exists some $R$-linear map $\alpha : F \to R$ such that $\alpha(g) \neq 0$).

But since $m_1 \wedge (rm_2) \wedge ... \wedge m_n = 0$, the alternating $n$-form $$x_1 f(x_2, ..., x_n) - x_2 f(x_1, x_3, ..., x_n) \pm ...$$ necessarily vanishes on it, so $$m_1 f(rm_2, ..., m_n) = m_1 s = rm_2 f(m_1, m_3, ..., m_n) \mp ...$$ and we conclude that the $m_i$ are linearly dependent.

Corollary: Let $f : M \to N$ be an injective map of free modules. Then the induced map $\Lambda(f) : \Lambda(M) \to \Lambda(N)$ is also injective.

Proof. Suppose otherwise. Let $e_i, i \in I$ be an ordered basis of $M$. For a finite subset $S$ of $I$, let $e_S$ denote the wedge of all of the elements of $S$ (in the order determined by the order of $I$). If $\Lambda(f)$ is not injective, then let $\sum c_S e_S$ be some element of its kernel. Since the kernel is a two-sided ideal (because $\Lambda(f)$ is a ring homomorphism), we may freely take exterior products on either side (this is also crucial!). Now, if $\sum c_S e_S$ has at least two nonzero terms in it, then there exists $i$ such that $e_i$ appears in some term but not every term, so it follows that $$\sum c_S e_S \wedge e_i \in \text{ker}(\Lambda(f))$$

as well. This is an element of the kernel with strictly fewer nonzero terms. Hence an element of the kernel with the minimal number of nonzero terms necessarily has a single nonzero term $c_S e_S$. Writing this as $$c_S e_1 \wedge ... \wedge e_k$$

and applying $\Lambda(f)$ we get $$c_S f(e_1) \wedge ... \wedge f(e_k) = 0.$$

By the above, it follows that $f(e_1), ... f(e_k)$ are linearly dependent, but this contradicts $f$ injective.

Corollary: Let $f : R^n \to R^n$ be an endomorphism of a free module. Then $f$ is injective if and only if $\det(f)$ is not a zero divisor.

Proof. By the above, if $f$ is injective then $\Lambda^n(f) : \Lambda^n(R^n) \to \Lambda^n(R^n)$ is also injective. Since it acts by multiplication by $\det(f)$, we conclude that $\det(f)$ is not a zero divisor. If $f$ is not injective then some $\lambda_1 e_1 + \lambda_2 e_2 + \cdots + \lambda_n e_n$ lies in the kernel of $\Lambda(f)$, where not all $\lambda_i$ are $0$; thus, by taking exterior products we conclude that $\lambda_i e_1 \wedge ... \wedge e_n$ also lies in the kernel of $\Lambda(f)$ for all $i$, and therefore $\lambda_i \det(f) = 0$ for all $i$.

Lam's Exercises in modules and rings includes the following:

which tells us that your determinant is not a zero-divisor.

The paper where McCoy does that is [Remarks on divisors of zero, MAA Monthly 49 (1942), 286--295] If you have JStor access, this is at http://www.jstor.org/stable/2303094

There is a pretty corollary there: a square matrix is a zero-divisor in the ring of matrices over a commmutative ring iff its determinant is a zero divisor.