Free modules
I think it should be true when $M,N$ are free. My proof is not complete yet, but here is some progress.
By using standard arguments with filtered colimits, we can easily deduce to the finite case. Then the question is just: If $f : R^m \to R^n$ is an injective linear map, represented by a matrix $(a_{ij}) \in R^{n \times m}$, is the induced algebra homomorphism
$$\mathrm{Sym}(f) : R[T_1,\dotsc,T_m] \to R[T_1,\dotsc,T_n], ~ T_j \mapsto \sum_i a_{ij} T_i$$
also injective? This is a homomorphism of graded algebras, so it suffices to look at homogeneous polynomials of a given degree. The degrees $0$ and $1$ are easy.
If $R$ is a field, it is of course true (see also Eric's answer). It then follows also when $R$ is an integral domain.
Let us consider the special case $m=1$, so $f$ corresponds to an element $(a_1,\dotsc,a_n) \in R^n$ with $\mathrm{Ann}(a_1,\dotsc,a_n)=0$, and $\mathrm{Sym}(f)$ maps $T_1 \mapsto \sum_i a_i T_i$. Here is a proof that $\mathrm{Sym}(f)$ is, in fact, injective:
Take an element $r T_1^d$ in the kernel (as said earlier, it suffices to consider homogeneous elements). By the multinomial theorem, this means
$$0 = r \sum_{i_1+\cdots + i_n=d} \binom{d}{i_1 ~ \cdots ~ i_n} a_1^{i_1} \cdots a_n^{i_n}.$$
for all partitions $i_1+\cdots+i_n=d$. In particular,
$$r \cdot a_i^d = 0$$
for all indices $i$. Let $s := n \cdot d$. By the pigeonhole principle, for all indices $i_1,\dotsc,i_s$ we deduce
$$r \cdot a_{i_1} \cdots a_{i_s} = 0,$$
since in fact every such product will contain some $a_i$ at least $d$ times. But then, since $\mathrm{Ann}(a_1,\dotsc,a_n)=0$, we deduce
$$r \cdot a_{i_1} \cdots a_{i_{s-1}} = 0$$
for all indices $i_1,\dotsc,i_{s-1}$. We can continue inductively in the same way to reduce the number of factors. In the end, no factors are left, so that $r = 0$ and we are done.
Flat modules in characteristic $0$
The result holds when $M,N$ are flat and $R$ is a $\mathbb{Q}$-algebra: The natural epimorphism $M^{\otimes d} \to \mathrm{Sym}^d(M)$ splits via $$\omega_M : \mathrm{Sym}^d(M) \to M^{\otimes d},~m_1 \cdots m_d \mapsto \frac{1}{d!} \sum_{\sigma \in \Sigma_d} m_{\sigma(1)} \otimes \cdot \otimes m_{\sigma(d)}.$$
If $f : M \to N$ is a linear map, the diagram
$$\require{AMScd}
\begin{CD} \mathrm{Sym}^d(M) @>{\mathrm{Sym}^d(f)}>> \mathrm{Sym}^d(N) \\ @V{\omega_M}VV @VV{\omega_N}V \\ M^{\otimes d} @>>{f^{\otimes d}}> N^{\otimes d} \end{CD}$$
commutes. If $f$ is injective, since $M$ and $N$ are flat, $f^{\otimes d}$ is injective. Since $\omega_M$ is injective (even a split monomorphism), we see that $\mathrm{Sym}^d(f)$ is injective .
So if $R$ is arbitrary (but $M,N$ are still flat), use the localization functor $M \mapsto M_{\mathbb{Q}}$ to deduce that $\mathrm{Sym}^d(f)_{\mathbb{Q}} = \mathrm{Sym}^d(f_{\mathbb{Q}})$ is injective. This means that any element in the kernel of $\mathrm{Sym}^d(f)$ is killed by some non-zero integer (in fact, by $d!$). Since in our proof above we see nothing about these integers, I highly suspect that $\mathrm{Sym}^d(f)$ is injective, but I haven't finished the proof.
