First, here is a heuristic explanation of why the answer is no. The hypersurface $\Gamma \subseteq \mathbb{C}^2$ is of real codimension two, and a homotopy $H: [0,1]^2 \to \mathbb{C}^2$ will also be of real codimension two (if we choose a smooth approximation, etc. etc.). Hence, we expect them to generically intersect in real codimension four, i.e. in a $0$-dimensional submanifold of $\mathbb{C}^2$. This need not be empty. This is exactly the phenomenon that leads $\mathbb{C}^2 \setminus \Gamma$ to often have interesting fundamental groups.
Here is a counterexample: let $\Gamma = (\mathbb{C}\times 0) \cup (\mathbb{C} \times 1) \cup (0 \times \mathbb{C})$, and let $X = \mathbb{C}^2 \setminus \Gamma$. Let $\gamma_1: [0,1] \to \mathbb{C}^2$ be a loop based at $(1,0)$ such that $\gamma$ does not wind around $\mathbb{C} \times 1$ or $0 \times \mathbb{C}$, and let $\gamma_2: [0,1] \to \mathbb{C}^2$ be a loop which winds around $\mathbb{C}\times 1$ once, e.g.
$$\gamma_1(t) = (1, a (1 - e^{2\pi i t})),$$
$$ \gamma_2(t) = (1, b (1-e^{2\pi i t}))$$
for $a < 1/2 < b$.
Figure: The loops depicted are homotopic to
a loop in the hypersurface, but not each other.
Both are partially homotopic in your sense to the loop $\gamma$ which travels linearly in $\Gamma$ from $(1,0)$ to $(0,0)$ to $(0,1)$ to $(1,1)$, and then back. But $\gamma_1$ and $\gamma_2$ are not partially homotopic in $X$. Consider $Y = \mathbb{C}^2 \setminus \mathbb{C} \times 1$. If $\gamma_1$ and $\gamma_2$ were partially homotopic in $X$, then they would be homotopic relative to the basepoint $(1,0)$ in $Y$, i.e. define the same class in $\pi_1(Y,(1,0))$. But they do not, since $[\gamma_2] \in \pi_1(Y,(1,0))$ is a generator, while $0 = [\gamma_1] \in \pi_1(Y,(1,0))$.
In the case that $n$ and $m$ are both even, this comes down to knowing that $\pi_1(S^1)\cong \mathbb{Z}$ is generated by the homotopy class of the path $\gamma(t) = (cos(2\pi t), sin(2\pi t))$. Based on this, a simple argument goes as follows:
Say $n = 2k$ and $m = 2l$. First, note that there are homotopies $f \sim k\gamma$ and $g \sim l \gamma$ (where $k\gamma$ is the concatenation of $k$ copies of $\gamma$). If $n\neq m$ then $k[\gamma]\neq l[\gamma]$ since $\pi_1(S^1)$ is freely generated by $[\gamma]$, so the result follows for $S^1$. The inclusion $S^1 \to \mathbb{R}^2\setminus\{0,0\}$ is a homotopy-equivalence (the homotopy-inverse sends $v$ to $\frac{v}{|v|}$) so it induces an isomorphism on $\pi_1$, therefore if $f\sim g$ in $\mathbb{R}^2\setminus\{0,0\}$ then they must be homotopic in $S^1$ as well, which is a contradiction.
The only other case is where $n$ and $m$ are both odd, so that $f$ and $g$ both start at $p = (1,0)$ but end at $q = (-1,0)$. This case is slightly weird since we're used to dealing with loops, but we will leverage the fact that there is a bijection between $\pi_1(S^1)$ and the set of homotopy classes of paths relative endpoints from $p$ to $q$, denoted $\pi_0(\Omega_{p,q}(S^1))$ (here $\Omega_{p,q}(S^1)$ denotes the space of paths from $p$ to $q$, similar to the notation for the loop space, and the path components of this space are the homotopy classes rel. endpoints).
One such a bijection is given by taking the path $\alpha$ going counterclockwise from $q$ to $p$, and sending the class of a path $[\beta]\in \pi_0(\Omega_{p,q}(S^1))$ to the class of the loop $[\beta * \alpha]\in \pi_1(S^1)$; the inverse is given by sending a loop $[\delta]$ to $[\delta* \alpha^{-1}]$.
Now we need to determine the loops given by concatenating $f$ and $g$ with $\alpha$ (I will give a sketch). If $n = 2k + 1 > 0$ then $f$ is a path which goes counterclockwise around the circle $k$ and a half times and so $f*\alpha$ goes around the circle $k + 1$ times, i.e. $[f * \alpha] = (k+1)[\gamma]$; if $n<0$ and hence $k = \frac{n-1}{2}< 0$ then $f$ goes clockwise around the circle $|k|$ minus a half times (draw some pictures for different values of $n$ and $k$), and contatenating with $\alpha$ backtracks a half circle and we get $[f*\alpha] = (|k| - 1)[\gamma^{-1}] = -(|k| - 1)[\gamma] = (k+1)[\gamma]$. Similarly if $m= 2l + 1$ then $[g * \alpha] = (l+1)[\gamma]$. Again since $\pi_1(S^1)$ is generated by $[\gamma]$ and the map $[\_*\alpha]$ is a bijection it follows that $f$ and $g$ are homotopic relative endpoints iff $n = m$.
For your concrete example $n = 1$ and $m = -1$, then $k = 0$ and $l = -1$ and we get $[f * \alpha] = [\gamma]$ and $[g* \alpha] = 0$. In fact $g\sim \alpha^{-1}$ in $\Omega_{p,q}(S^1)$.
Best Answer
Hint: any path $\gamma$, open or closed, is homotopic to the constant path $t \mapsto \gamma(1/2)$. Any constant path is homotopic to any other constant path in the same path component.